1. Euclid's division lemma states that for any two positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b.
2. If a divides b and b divides a, then a = b or a = -b. Since we are typically dealing with positive integers in this context, a = b.
3. 0.3 as a fraction in simplest form is 3/10.
4. The rational number 17/125 is terminating because the denominator 125 = 5³ has only prime factor 5.
5. A rational number p/q is a terminating decimal if and only if the prime factorization of q (in lowest terms) contains only the primes 2 and/or 5.
6. For any integer a and divisor 3, the possible values of remainder r are 0, 1, or 2.
7. HCF(480, 672) = 96, since LCM × HCF = product of numbers → 3360 × HCF = 480 × 672 → HCF = (480 × 672)/3360 = 96.
8. The number is 117 × 41 + 23 = 4797 + 23 = 4820.
9. Given p = ab² and q = a³b, where a and b are prime, LCM(p, q) = a³b².
10. Given a = (2⁴ × 3¹ × 5⁶) and b = (2³ × 3⁵ × 5), HCF(a, b) = 2³ × 3¹ × 5¹ = 8 × 3 × 5 = 120.
11. HCF(65, 117) = 13. Since 65m - 117 = 13, then 65m = 130 → m = 2.
12. x = 2.23̅ = 2.2333... is rational because it is a repeating decimal, which can be expressed as a fraction: let x = 2.2333..., then 10x = 22.333... and 100x = 223.333..., so 90x = 201 → x = 201/90 = 67/30.
13. Let HCF = h, then LCM = 45h. Given one number is 125, and h + 45h = 1150 → 46h = 1150 → h = 25. So LCM = 45×25 = 1125. Since LCM × HCF = product of numbers, 1125 × 25 = 125 × other number → other number = (1125 × 25)/125 = 225.
14. Using Euclid’s algorithm for 867 and 255:
867 = 3 × 255 + 102
255 = 2 × 102 + 51
102 = 2 × 51 + 0
So HCF = 51.
15. 21ⁿ ends with 0, 2, 4, 6, or 8 only if it is even. But 21 is odd, so 21ⁿ is always odd for any natural number n. Therefore, it never ends with 0, 2, 4, 6, or 8.
16. Let the divisor be d. Then 1365 = 31d + 32 → 31d = 1333 → d = 1333 ÷ 31 = 43.
Parent Tip: Review the logic above to help your child master the concept of the real number math worksheet.
