Number System — PPSC previous year questions

The product of the two digits must be 4, 9, 16, 25, 36, 49, 64 or 81 (a perfect square bigger than 1, so numbers with a 0 or product 1 are out). > product 4: 14, 41, 22 (3) > product 9: 19, 91, 33 (3) > product 16: 28, 82, 44 (3) > product 36: 49, 94, 66 (3) > products 25, 49, 64, 81: 55, 77, 88, 99 (4) That gives 3 + 3 + 3 + 3 + 4 = 16 numbers.

Let the tens digit be a and the units digit be b. So a + b = 9. Taking away 45 swaps the digits. So (10a + b) − 45 = 10b + a. > 9(a − b) = 45 → a − b = 5 > a + b = 9 → a = 7, b = 2 The number is 72. Check: 72 − 45 = 27.

Two successive numbers n and n + 1 multiply to 1980. Since 44² = 1936 and 45² = 2025, the numbers lie around 44–45. > 44 × 45 = 1980 The smaller number is 44.

Writing a three-digit number twice multiplies it by 1001, because ABCABC = ABC × 1000 + ABC. > ABC × DEED = ABC × 1001 → DEED = 1001 So D = 1 and E = 0.

'Not both odd' means at least one of p and q is even. A product with an even factor is always even. So pq must be even. The others can fail. Take p = 2 and q = 3: p − q = −1 and p + q = 5 are both odd. And 2(p + q) − 1 is always odd for any integers.

Since q = 3p + 2, dividing q by 3 always leaves remainder 2. So q can never be a multiple of 3. The other options are possible for suitable p. p = 2 gives q = 8, which 4 divides. p = 1 gives q = 5. p = 4 gives q = 14, which 7 divides.

Only the unit digits of the bases matter. Powers of 4 cycle as 4, 6, 4, 6 …, so an odd power such as 4¹⁷⁹³ ends in 4; every power of 5 ends in 5; and 1678491 ends in 1. > 4 × 5 × 1 = 20 The product ends in 0. Whenever a factor ending in an even digit is multiplied by a factor ending in 5 the unit digit is always 0.

Write the number as X = 10a + b. > (10b + a) − (10a + b) = 63 gives 9(b − a) = 63, so b − a = 7 > With a + b = 11, a = 2 and b = 9, so X = 29 > X + 21 = 50

Let the number be n and the prime be p < n. The condition is n = n² − np. > n² − np = n → n(n − p) = n → n − p = 1 → n = p + 1 An awesome number is one more than a prime. For 10 < n < 99 the primes p from 11 to 97 are: 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. That is 21 primes, so 21 awesome numbers (12 to 98). So 'more than 16 but less than 22'.

Count the digits from 1 to 100. > '9': units place 9, 19, ..., 99 = 10; tens place 90–99 = 10 → 20 times (not 19) > '0': 10, 20, ..., 90 = 9, plus two zeros in 100 → 11 times (not 10) > '1': units 1, 11, ..., 91 = 10; tens 10–19 = 10; hundreds in 100 = 1 → 21 times ≠ 11 So (a), (b) and (d) are all false. In words, none of one, two, ..., ninety-nine or one hundred contains the letter 'a'. The first number word with an 'a' is 'thousand'.