8-Digit Number Challenges worksheet for practicing number formation and arithmetic skills.

Let’s go step by step through each question.

We are given the digits: 0, 1, 2, 3, 5, 6, 7, 9 — that’s 8 digits. We must use ALL of them exactly once to make an 8-digit number for each question (unless otherwise specified).

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1) What is the largest 8-digit number you can make?

To make the largest number, we arrange the digits in descending order:
→ 9, 7, 6, 5, 3, 2, 1, 0
So: 97,653,210

✔ Check: All digits used? Yes. Largest possible? Yes.

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2) Subtract 900 from this number.

97,653,210 – 900 = ?

Let’s compute:

97,653,210
– 900
= 97,652,310

✔ Correct.

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3) What is the smallest 8-digit number you can make? (Your number cannot start with a zero!)

Smallest number → arrange digits in ascending order, but first digit can’t be 0.

So, smallest non-zero digit is 1 → put 1 first.

Then remaining digits in ascending order: 0, 2, 3, 5, 6, 7, 9

So: 10,235,679

✔ Check: Starts with 1? Yes. Smallest possible? Yes.

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4) Add 90,000 to this number.

10,235,679 + 90,000 = ?

10,235,679
+ 90,000
= 10,325,679

✔ Correct.

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5) What is the largest multiple of 50 you can make?

A multiple of 50 must end in 00, 50, or 00 — actually, more precisely: divisible by 50 → ends in 00 or 50.

But we only have one ‘0’ and one ‘5’. So we can’t make “00”. So it must end in 50.

So last two digits: 50

Now, to make the largest number, we want the biggest digits at the front.

Remaining digits: 9, 7, 6, 3, 2, 1 (we used 5 and 0)

Arrange them in descending order: 9, 7, 6, 3, 2, 1 → so first six digits: 976321

Then add 50 at the end → 97,632,150

Check: Ends in 50? Yes. Divisible by 50? Yes (since 50 ÷ 50 = 1). Largest possible? Yes, because we put largest digits first.

✔ Correct.

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6) What is the smallest even number you can make?

Even number → must end in an even digit: 0, 2, 6 (from our digits: 0,1,2,3,5,6,7,9 → evens are 0,2,6)

To make smallest number, we want smallest digits at front, but can’t start with 0.

Also, to make it even, last digit must be even.

Strategy: Make the number as small as possible → start with smallest non-zero digit (1), then arrange rest in ascending order, BUT ensure last digit is even.

Try ending with 0 → then first digit 1, then 2,3,5,6,7,9 → 12,356,790 → but wait, we can do better.

Actually, to minimize the number, we want smaller digits earlier, and the last digit should be the smallest possible even digit that allows the rest to be arranged minimally.

Best approach: Fix last digit as smallest even digit → 0.

Then arrange remaining digits (1,2,3,5,6,7,9) in ascending order → 1,2,3,5,6,7,9 → so number: 12,356,790

Is there a smaller one if we end with 2?

If last digit is 2, then first digit must be 1 (smallest non-zero), then remaining: 0,3,5,6,7,9 → arrange ascending: 0,3,5,6,7,9 → so 10,356,792 → which is smaller than 12,356,790!

Compare:

- End with 0: 12,356,790
- End with 2: 10,356,792 ← smaller!

What about ending with 6? Last digit 6 → first digit 1, then 0,2,3,5,7,9 → 10,235,796 → even smaller!

Wait, let’s compare:

End with 6: 10,235,796
End with 2: 10,356,792 → 10,235,796 is smaller.

End with 0: 12,356,790 → bigger.

So best is to end with 6? But 6 is not the smallest even digit — 0 and 2 are smaller.

Wait — we need to minimize the entire number, not just the last digit.

Actually, the key is: to make the smallest number, we want the leftmost digits to be as small as possible.

So:

Option 1: Last digit = 0 → then first digit = 1, then 2,3,5,6,7,9 → 12,356,790

Option 2: Last digit = 2 → first digit = 1, then 0,3,5,6,7,9 → 10,356,792

Option 3: Last digit = 6 → first digit = 1, then 0,2,3,5,7,9 → 10,235,796

Option 4: Last digit = 0 → but can we put 0 later? No, because if we end with 0, we can't use 0 in middle to reduce value.

Actually, Option 3: 10,235,796 is smaller than Option 2: 10,356,792

Is there a way to get even smaller?

What if we end with 0, but put 0 not at end? No — for even number, last digit must be even.

Another idea: Can we end with 0 and still get a smaller number? Only if we can put smaller digits earlier.

But if we end with 0, then the first digit must be 1, and the next smallest available is 2 — so 12... is forced.

Whereas if we end with 6, we can put 0 in the second position: 10...

So 10,235,796 is smaller than 12,356,790.

Can we end with 2 and get 10,235,796? No, because if we end with 2, we can't use 2 in the middle.

Wait — let's list all possibilities systematically.

We want the smallest 8-digit number using all digits, ending in even digit.

Start with smallest possible first digit: 1

Then second digit: smallest available → 0

Then third: 2

Fourth: 3

Fifth: 5

Sixth: 7

Seventh: 9

Eighth: must be even → but we've used 0,2,6? Wait, we haven't used 6 yet.

Digits: 0,1,2,3,5,6,7,9

If we fix first seven digits as 1,0,2,3,5,7,9 → then last digit must be 6 → which is even → perfect.

So number: 10,235,796

Is this valid? Uses all digits? 1,0,2,3,5,7,9,6 → yes.

Ends in 6 → even → yes.

Is there a smaller one?

Suppose we try to make the seventh digit smaller? But 9 is the only one left if we use 1,0,2,3,5,7 — no, 6 is smaller than 9, but we need to save an even digit for the end.

Alternative: What if we end with 0? Then we can't use 0 in the middle.

So: first digit 1, then 2,3,5,6,7,9,0 → 12,356,790 → larger than 10,235,796.

What if we end with 2? Then we can't use 2 in the middle.

First digit 1, then 0,3,5,6,7,9,2 → 10,356,792 → which is larger than 10,235,796.

What if we end with 6? As above: 10,235,796

Can we make it smaller by rearranging? For example, after 10, can we put 2, then 3, then 5, then 6? But 6 is needed for the end.

No — we have to save 6 for the end.

After 10, the next smallest digits are 2,3,5,7,9 — and we must end with 6.

So 10,235,796 is indeed the smallest.

But wait — what if we end with 0? And put 0 at the end, but use 2,3,5,6,7,9 in order — but then first digit 1, then 2,3,5,6,7,9,0 → 12,356,790 — bigger.

Or, can we put 0 not at start but later? For example, 10,235,679 — but that ends with 9 — odd — not even.

To make it even, last digit must be even.

So 10,235,796 is the smallest.

But let me confirm: is 10,235,796 even? Yes, ends with 6.

And uses all digits: 1,0,2,3,5,7,9,6 — yes.

Is there a number like 10,235,679? But that ends with 9 — odd.

Or 10,235,697 — ends with 7 — odd.

So yes, 10,235,796 is the smallest even number.

But wait — what about 10,235,679? Not even.

Another candidate: 10,235,697 — not even.

What if we end with 0? 12,356,790 — larger.

Or 10,235,796 vs 10,235,679 — but 679 ends with 9 — not even.

So yes, 10,235,796 is correct.

But earlier I thought of ending with 6, but is there a way to end with a smaller even digit and still get a smaller number?

For example, if we end with 2, we get 10,356,792 — which is larger than 10,235,796.

Similarly, ending with 0 gives 12,356,790 — larger.

So 10,235,796 is the smallest even number.

✔ Confirmed.

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7) What is the closest number to 40 million you can make?

40 million = 40,000,000

We need to make an 8-digit number using digits 0,1,2,3,5,6,7,9 that is closest to 40,000,000.

First, note: 40,000,000 has 8 digits, so our number will be around there.

The first digit should be as close to 4 as possible.

Available digits: 0,1,2,3,5,6,7,9 — no 4.

So closest first digits: 3 or 5.

If we start with 3, the number is about 30 million — too low.

If we start with 5, it's 50 million — too high.

But 30 million is 10 million away, 50 million is 10 million away — same distance? But we can get closer by choosing the right digits.

Actually, 40 million is 40,000,000.

Numbers starting with 3: max is 39,765,210 — which is 40,000,000 - 39,765,210 = 234,790 away.

Numbers starting with 5: min is 50,123,679 — which is 50,123,679 - 40,000,000 = 10,123,679 away — much farther.

So better to start with 3.

Max number starting with 3: arrange remaining digits in descending order: 9,7,6,5,2,1,0 → so 39,765,210

Difference: 40,000,000 - 39,765,210 = 234,790

Is there a number starting with 3 that is closer? Probably not, since 39,765,210 is the largest possible starting with 3.

What about starting with 4? But we don't have 4.

Or starting with other digits? 2 would be even smaller.

So 39,765,210 is the closest? But let's check if we can get closer by not taking the maximum.

For example, if we take 39,765,210 — difference 234,790

Can we make a number like 39,765,201? But that's smaller — farther from 40M.

Or 39,765,210 is the largest, so closest from below.

From above, the smallest starting with 5 is 50,123,679 — difference over 10 million.

But what about starting with 3, but making it larger? We can't — 39,765,210 is max.

Unless... is there a number starting with 3 that is greater than 39,765,210? No, because we used the largest digits after 3.

But 40 million is 40,000,000, and 39,765,210 is less, but perhaps we can make a number like 39,999,999 — but we don't have three 9's.

We have only one 9.

So 39,765,210 is the best we can do starting with 3.

But let's calculate the difference: 40,000,000 - 39,765,210 = 234,790

Is there a number starting with 3 that is closer? For example, if we could make 39,999,999, but we can't.

Perhaps starting with 4 is impossible.

Another idea: what if we start with 3, but arrange the digits to be as large as possible, which we did.

But let's see: is 39,765,210 the closest? Or can we make a number like 39,765,201? That's smaller — farther.

Or 39,765,210 is fixed.

But wait — 40 million is 40,000,000, and we have digits including 9,7,6,5, etc.

Perhaps we can make 39,765,210, but let's see if there's a number closer to 40M.

What about 39,765,210 vs 39,765,210 — same.

Or perhaps 39,765,210 is not the only option; maybe we can make 39,765,210, but let's calculate the actual difference.

40,000,000 - 39,765,210 = 234,790

Now, is there a number starting with 3 that is larger than this? No.

What about starting with 4? Not possible.

Starting with 5: minimum is 50,123,679 — difference 10,123,679 — much larger.

So 39,765,210 is the closest.

But let's double-check: can we make 39,765,210? Digits used: 3,9,7,6,5,2,1,0 — yes, all digits.

And it's 39,765,210.

Difference to 40,000,000 is 234,790.

Is there a number like 39,765,210 — same.

Or perhaps 39,765,201 — but that's 39,765,201, difference 234,799 — larger difference.

So yes, 39,765,210 is closest.

But wait — what if we make 39,765,210, but is there a number like 39,765,210 — I think it's correct.

Another thought: 40 million is 40,000,000, and we have digit 4? No.

Perhaps we can make 39,999,999, but we don't have multiple 9's.

So yes, 39,765,210 is the closest.

But let's confirm the calculation: 40,000,000 - 39,765,210 = 234,790 — yes.

And no other combination gets closer.

✔ So answer is 39,765,210

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8) Make an 8-digit number which is divisible by 4.

Divisible by 4: last two digits form a number divisible by 4.

So we need to choose last two digits from our digits such that the two-digit number is divisible by 4, and use all digits.

Possible pairs from {0,1,2,3,5,6,7,9} that form a number divisible by 4:

List all possible two-digit combinations from these digits (without repetition) that are divisible by 4.

Divisible by 4: 00,04,08,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,88,92,96

From our digits, possible:

- 00 — not possible, only one 0
- 04 — no 4
- 08 — no 8
- 12 — yes, digits 1,2
- 16 — yes
- 20 — yes
- 24 — no 4
- 28 — no 8
- 32 — yes
- 36 — yes
- 40 — no 4
- 44 — no
- 48 — no
- 52 — yes
- 56 — yes
- 60 — yes
- 64 — no 4
- 68 — no 8
- 72 — yes
- 76 — yes
- 80 — no 8
- 84 — no
- 88 — no
- 92 — yes
- 96 — yes

So possible last two digits: 12,16,20,32,36,52,56,60,72,76,92,96

Now, for each, we can make a number by arranging the remaining digits in any order, but since the question just says "make an 8-digit number", we can choose any.

To keep it simple, let's pick one, say 12.

Last two digits: 12

Remaining digits: 0,3,5,6,7,9

Arrange them in ascending order for simplicity: 0,3,5,6,7,9 → so first six digits: 035679 — but can't start with 0.

So arrange as 3,0,5,6,7,9 → 305679

Then add 12 at end: 305,679,12 — but that's 8 digits? 30567912 — yes.

Check: digits used: 3,0,5,6,7,9,1,2 — all present.

Last two digits 12 — 12 ÷ 4 = 3 — divisible by 4. Good.

We could choose other endings, but this works.

To make it larger or smaller, but not required.

So one possible answer: 30,567,912

But let's make sure it's valid.

Digits: 3,0,5,6,7,9,1,2 — yes, all from the set.

Ends with 12 — divisible by 4.

Perfect.

We could also choose 20: last two digits 20.

Remaining: 1,3,5,6,7,9 — arrange as 1,3,5,6,7,9 → 13567920 — also good.

But since the question doesn't specify, any is fine.

I'll go with 30,567,912

But to be safe, let's choose one that is straightforward.

Another common choice: 00 not possible, 20 is good.

Let me use 20.

Last two digits: 20

Remaining digits: 1,3,5,6,7,9

Arrange in ascending order: 1,3,5,6,7,9 → so number: 13,567,920

Check: digits 1,3,5,6,7,9,2,0 — all present.

Ends with 20 — 20 ÷ 4 = 5 — divisible by 4.

Good.

So 13,567,920

This is also fine.

I'll use this one.

✔ Answer: 13,567,920

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9) Make an 8-digit number which is divisible by 6.

Divisible by 6 means divisible by 2 and by 3.

Divisible by 2: last digit even.

Divisible by 3: sum of digits divisible by 3.

First, sum of all digits: 0+1+2+3+5+6+7+9 = let's calculate: 0+1=1, +2=3, +3=6, +5=11, +6=17, +7=24, +9=33.

33 is divisible by 3 (33÷3=11), so any number made from these digits will have digit sum 33, which is divisible by 3. So every such number is divisible by 3.

Therefore, to be divisible by 6, we just need it to be even — i.e., last digit even.

So any even number made from these digits will work.

From question 6, we have 10,235,796 which is even, so it should be divisible by 6.

Let me verify: sum of digits 33, divisible by 3; last digit 6, even — so yes, divisible by 6.

So we can use 10,235,796

Or any other even number.

To be consistent, I'll use the same as question 6.

✔ Answer: 10,235,796

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10) Write down 5 different numbers between 32,976,000 and 33,000,000 that you can make.

Range: greater than 32,976,000 and less than 33,000,000.

So numbers from 32,976,001 to 32,999,999.

We need to make 8-digit numbers using digits 0,1,2,3,5,6,7,9.

First, the number must start with 32, because 33,000,000 is the upper limit, and 32,xxx,xxx is within range if xxx > 976,000.

Specifically, since lower bound is 32,976,000, so the number must be at least 32,976,001.

So first two digits: 3 and 2.

Then the next digits must make the number greater than 32,976,000.

So after 32, the next three digits should be at least 976.

But we have to use the remaining digits: 0,1,5,6,7,9 (since 3 and 2 are used).

Digits available: 0,1,5,6,7,9

We need to form the last six digits such that the number is between 32,976,000 and 33,000,000.

Since it starts with 32, and we need it > 32,976,000, so the number formed by digits 3 to 8 should be > 976,000.

That is, the first digit after 32 should be 9 (since 976,000 starts with 9), and then the next digits should be such that it's at least 976,000.

So, position 3 (hundred thousands place) must be 9, because if it's less than 9, say 7, then 32,7xx,xxx < 32,976,000.

Similarly, if it's 9, then we need the next digits to be at least 76,000.

So, fix first three digits: 3,2,9

Then remaining digits: 0,1,5,6,7 (since 3,2,9 used)

Now, the number is 32,9xx,xxx

We need it > 32,976,000, so the number formed by the last five digits must be > 76,000.

That is, the next digit (ten thousands place) should be at least 7.

Available digits: 0,1,5,6,7

So, if we put 7 in ten thousands place, then we need the remaining to be > 6,000? Let's see.

The number is 32,9ab,cde

We need 32,9ab,cde > 32,976,000

Since first three digits are 32,9, we compare the next part: ab,cde > 76,000

ab,cde is a 5-digit number.

So ab,cde > 76,000

With digits available: 0,1,5,6,7

To make ab,cde > 76,000, the first digit a should be at least 7.

If a=7, then b,cde > 6,000

Available digits after using 7: 0,1,5,6

So b should be at least 6? 6,000 is the threshold.

If b=6, then cde > 000, which is always true since digits are positive, but cde could be 000, but we have to use all digits, and 0,1,5 left, so minimum cde is 015 or something, which is >0.

But we need b,cde > 6,000? When a=7, we need the number 7b,cde > 76,000, so b,cde > 6,000.

Since b is the thousands digit, if b>=6, then it's at least 6,000, but we need >6,000? 76,000 is the threshold, so if b=6, then 76,000 is equal, but we need greater than 32,976,000, so strictly greater.

32,976,000 is the lower bound, and we need numbers greater than that, so 32,976,001 and above.

So if we make 32,976,xxx, it could be equal or greater.

But 32,976,000 is not included, since it says "between", and typically in such contexts, it might be exclusive, but let's check the problem: "between 32,976,000 and 33,000,000" — usually means exclusive, but sometimes inclusive. To be safe, let's assume exclusive, so greater than 32,976,000 and less than 33,000,000.

So 32,976,000 is not included.

So we need the number > 32,976,000.

So if we have 32,976,xxx, it must be > 32,976,000, so xxx > 000, which is true as long as not all zero, but we have digits to use.

But in our case, with digits 0,1,5,6,7, we can make various numbers.

Back to construction.

First three digits: 3,2,9

Remaining digits: 0,1,5,6,7

We need the number 32,9ab,cde > 32,976,000

So ab,cde > 76,000

ab,cde is a 5-digit number formed by digits 0,1,5,6,7.

To make it > 76,000, the first digit a should be 7 (since if a<7, say 6, then max is 67,510 < 76,000? 67,510 < 76,000 yes, so a must be 7.

So a=7

Then b,cde > 6,000

Available digits: 0,1,5,6

b should be at least 6? If b=6, then cde > 000, which is true.

If b<6, say 5, then 5cde < 6,000? 5,xxx < 6,000, so not >6,000.

So b must be 6.

Then cde > 000, which is always true with remaining digits 0,1,5.

So we can arrange c,d,e as 0,1,5 in any order.

So possible numbers: 32,976,015; 32,976,051; 32,976,105; etc.

All these are greater than 32,976,000 and less than 33,000,000.

Also, we could have other combinations, but this is one way.

For example, if we put a=7, b=6, then c,d,e from 0,1,5.

So five different numbers:

1. 32,976,015

2. 32,976,051

3. 32,976,105

4. 32,976,150

5. 32,976,501

All use digits: 3,2,9,7,6, and then 0,1,5 in different orders.

Check digits: for 32,976,015: digits 3,2,9,7,6,0,1,5 — all present.

Similarly for others.

And all are between 32,976,000 and 33,000,000.

Perfect.

So five numbers: 32,976,015; 32,976,051; 32,976,105; 32,976,150; 32,976,501

✔ Good.

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11) Write these 5 numbers in order from smallest to largest.

The five numbers from question 10:

32,976,015

32,976,051

32,976,105

32,976,150

32,976,501

Now, sort them:

All have same first six digits: 32,976

Then last three digits: 015, 051, 105, 150, 501

So order: 015 < 051 < 105 < 150 < 501

So:

1. 32,976,015

2. 32,976,051

3. 32,976,105

4. 32,976,150

5. 32,976,501

✔ Correct.

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12) Look at the number 63,107,295

Round it to various places.

First, the number: 63,107,295

a) Round to nearest 10.

Look at units digit: 5 → since 5>=5, round up.

Tens digit is 9, so 9+1=10, so carry over.

63,107,295 → units 5, so round up tens: 9 becomes 10, so tens become 0, hundreds increase by 1.

Hundreds is 2, becomes 3.

So 63,107,300

Confirm: 63,107,295 to nearest 10: since 5, round up, so 63,107,300

Yes.

b) Round to nearest 100.

Look at tens digit: 9, which is >=5, so round up hundreds.

Hundreds digit is 2, becomes 3.

Tens and units become 00.

So 63,107,300

Same as above? Let's see.

Number: 63,107,295

To nearest 100: look at tens digit: 9 >=5, so round up hundreds.

Hundreds is 2, so 2+1=3, and set tens and units to 00.

So 63,107,300

Yes.

c) Round to nearest 1000.

Look at hundreds digit: 2, which is <5, so round down.

So thousands and above stay, hundreds, tens, units become 000.

So 63,107,000

Number is 63,107,295

Thousands digit is 7 (since 107,295 — the 7 is thousands place? Let's write with commas: 63,107,295

So:

- Millions: 63

- Thousands: 107

- Units: 295

So to nearest 1000: look at the hundreds digit of the whole number, which is the digit in hundreds place: 2 (from 295)

Since 2<5, we round down, so keep thousands as is, set last three digits to 000.

So 63,107,000

Yes.

d) Round to nearest 10,000.

Look at thousands digit: 7 (from 107,295 — the 7 is thousands? 107,295 means 107 thousand, so thousands digit is 7.

Standard: 63,107,295

Positions:

- Ten millions: 6

- Millions: 3

- Hundred thousands: 1

- Ten thousands: 0

- Thousands: 7

- Hundreds: 2

- Tens: 9

- Units: 5

To round to nearest 10,000, look at the thousands digit (which is 7).

Since 7 >=5, we round up the ten thousands digit.

Ten thousands digit is 0, so 0+1=1, and set thousands and below to 0000.

So the number becomes 63,110,000

Because: 63,107,295 → ten thousands is 0, thousands is 7>=5, so round up ten thousands: 0 to 1, and lower digits 0.

So 63,110,000

Yes.

e) Round to nearest 100,000.

Look at ten thousands digit: 0 (from above)

0 < 5, so round down.

So hundred thousands digit stays 1, and lower digits become 00,000.

So 63,100,000

Number: 63,107,295

Hundred thousands digit is 1 (since 107,295 — 1 is hundred thousands)

Ten thousands digit is 0 <5, so round down, so 63,100,000

Yes.

f) Round to nearest million.

Look at hundred thousands digit: 1 <5, so round down.

Millions digit is 3, so keep 63,000,000

Number: 63,107,295

Millions: 63 million

Hundred thousands: 1 <5, so round down to 63,000,000

Yes.

So summary:

- Nearest 10: 63,107,300

- Nearest 100: 63,107,300

- Nearest 1000: 63,107,000

- Nearest 10,000: 63,110,000

- Nearest 100,000: 63,100,000

- Nearest million: 63,000,000

✔ All correct.

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Now, compiling all answers:

1) 97,653,210

2) 97,652,310

3) 10,235,679

4) 10,325,679

5) 97,632,150

6) 10,235,796

7) 39,765,210

8) 13,567,920 (or any valid, but I'll use this)

9) 10,235,796

10) 32,976,015; 32,976,051; 32,976,105; 32,976,150; 32,976,501

11) 32,976,015; 32,976,051; 32,976,105; 32,976,150; 32,976,501

12)
- Nearest 10: 63,107,300
- Nearest 100: 63,107,300
- Nearest 1000: 63,107,000
- Nearest 10,000: 63,110,000
- Nearest 100,000: 63,100,000
- Nearest million: 63,000,000

For question 8, I used 13,567,920, but let me confirm if it's divisible by 4: last two digits 20, 20÷4=5, yes.

And digits: 1,3,5,6,7,9,2,0 — all present.

Good.

For question 9, I used 10,235,796, which is even and sum 33 divisible by 3, so divisible by 6.

Good.

Now, for the final answer, since the user didn't specify to explain, and options are false, I just need to provide the final answers in order.

But the response format says to provide final answer after reasoning, and since no explanation or teach or practice is requested, I'll just list the answers.

But in the final answer section, I should put all the answers clearly.

Since it's a worksheet, I'll list them as per the questions.

Final Answer:

1) 97653210
2) 97652310
3) 10235679
4) 10325679
5) 97632150
6) 10235796
7) 39765210
8) 13567920
9) 10235796
10) 32976015, 32976051, 32976105, 32976150, 32976501
11) 32976015, 32976051, 32976105, 32976150, 32976501
12)
- Nearest 10: 63107300
- Nearest 100: 63107300
- Nearest 1000: 63107000
- Nearest 10000: 63110000
- Nearest 100000: 63100000
- Nearest million: 63000000

I'll write them with commas for clarity, but in the box, perhaps without commas, but since it's text, I can use commas.

In the initial, they used commas, so I'll keep commas.

But in the final answer, to be precise, I'll write as numbers.

Since it's a text response, I'll format it nicely.

For question 10 and 11, I need to list five numbers.

In question 12, list the rounded values.

So, final answer section: