Number system and divisibility questions, with answers

The questions, with answers

1. 1.What are the divisibility rules you need to know, and why do they work?

   By 2, 5 and 10: look at the last digit. By 4 and 25: the last two digits; by 8 and 125: the last three, because 100 is divisible by 4 and 25, and 1000 by 8 and 125, so only the tail matters. By 3 and 9: the digit sum, because 10, 100 and 1000 all leave remainder 1 when divided by 3 or 9, so a number and its digit sum leave the same remainder. By 11: the alternating sum of digits (units minus tens plus hundreds ...) must be 0 or a multiple of 11, because 10 leaves remainder -1 when divided by 11. By 6: both 2 and 3; by 12: both 3 and 4 — always a pair of coprime factors, never 2 and 6. For 7 and 13 there are rules, but the test rarely wants them; long division is faster.

2. 2.How do you test a number for divisibility by 11?

   Add the digits in the odd positions and the digits in the even positions, counting from the right, and subtract one total from the other; if the difference is 0 or divisible by 11, so is the number. For 2,728: from the right, 8 + 7 = 15 and 2 + 2 = 4, difference 11, so 2,728 = 11 × 248 is divisible. For 9,174: 4 + 1 = 5 and 7 + 9 = 16, difference 11, divisible (11 × 834). The rule works because every even power of 10 leaves remainder 1 on division by 11 and every odd power leaves remainder 10, which is -1. Questions often hand you a number with a missing digit and ask what makes it divisible by 11 — set the alternating sum to 0 or ±11 and solve for the digit.

3. 3.How do you find the units digit of a large power?

   Units digits of powers cycle, and the cycle depends only on the base's own units digit. Digits 0, 1, 5 and 6 never change: 6 to any power ends in 6. Digits 4 and 9 alternate with period 2: 4, 6, 4, 6 and 9, 1, 9, 1. Digits 2, 3, 7 and 8 have period 4: 2 goes 2, 4, 8, 6; 3 goes 3, 9, 7, 1; 7 goes 7, 9, 3, 1; 8 goes 8, 4, 2, 6. So divide the exponent by the period and use the remainder, treating remainder 0 as the last entry of the cycle. Units digit of 3^23: 23 ÷ 4 leaves remainder 3, the third entry in 3, 9, 7, 1 is 7. Units digit of 8^40: remainder 0, last entry, 6. Two-digit bases use only their last digit, so 27^15 behaves like 7^15.

4. 4.How do you find the remainder of a large power, like 2^50 divided by 7?

   Find a power of the base that leaves remainder 1, then reduce the exponent. 2^3 = 8 leaves remainder 1 when divided by 7, so 2^50 = (2^3)^16 × 2^2, which leaves the same remainder as 1^16 × 4 = 4. In general, remainders of successive powers repeat in a cycle, and once you find the cycle length you only need the exponent modulo that length. The same idea handles a base larger than the divisor: replace the base by its remainder first, so 100^7 divided by 9 becomes 1^7 = 1, because 100 leaves remainder 1 on division by 9. Fermat's little theorem gives the cycle length for a prime divisor — a^(p-1) leaves remainder 1 — but on a placement test spotting the small power that gives 1 is quicker.

5. 5.What are HCF and LCM, and what is the relation between them?

   The HCF (GCD) of two numbers is the largest number that divides both; the LCM is the smallest number both divide. For two numbers, HCF × LCM = product of the numbers, which lets you find one from the other: if the HCF of two numbers is 6 and their product is 720, the LCM is 120. The relation does not extend to three numbers. To compute them, factorise: 36 = 2² × 3² and 48 = 2⁴ × 3, so the HCF takes the lower power of each prime, 2² × 3 = 12, and the LCM the higher, 2⁴ × 3² = 144; check 12 × 144 = 1,728 = 36 × 48. For big numbers, Euclid's algorithm — replace the larger by its remainder on division by the smaller until one is zero — finds the HCF in a few steps.

6. 6.How do you solve "the smallest number that leaves the same remainder when divided by several numbers"?

   The number is a multiple of the LCM of the divisors, plus the remainder. The smallest number leaving remainder 2 when divided by 4, 6 and 10: LCM(4, 6, 10) = 60, so the answer is 62. With different remainders whose shortfalls are equal — leaving 1 when divided by 4, 3 when divided by 6, 7 when divided by 10, each 3 short of the divisor — the number is a multiple of the LCM minus that common shortfall: 60 - 3 = 57. The mirror question asks for the largest number that divides several numbers leaving given remainders: subtract each remainder from its number and take the HCF of the results. So the largest number dividing 62 and 95 leaving remainders 2 and 5 is HCF(60, 90) = 30.

7. 7.How do you count the factors of a number, and find their sum?

   From the prime factorisation. If n = p^a × q^b × r^c, the number of factors is (a + 1)(b + 1)(c + 1), because a factor chooses an exponent for each prime from 0 up to its maximum. 360 = 2³ × 3² × 5 has (3 + 1)(2 + 1)(1 + 1) = 24 factors. The sum of the factors is the product of the geometric series for each prime: (1 + 2 + 4 + 8)(1 + 3 + 9)(1 + 5) = 15 × 13 × 6 = 1,170. Related asks: a number has an odd number of factors exactly when it is a perfect square, since only then does a factor pair up with itself; and the number of ways to write n as a product of two factors is half the factor count, rounded up for squares.

8. 8.How many trailing zeros does a factorial have?

   Count the factors of 5, because every trailing zero is a factor of 10 = 2 × 5 and factorials always have far more 2s than 5s. Add the quotients n ÷ 5, n ÷ 25, n ÷ 125 and so on, taking only the whole-number part each time, until the quotient is 0. For 100!: 100 ÷ 5 = 20, 100 ÷ 25 = 4, 100 ÷ 125 = 0, so 24 trailing zeros — the 25 term matters because multiples of 25 contribute two 5s each. For 27!: 27 ÷ 5 = 5, 27 ÷ 25 = 1, total 6. The same technique gives the highest power of any prime that divides n!, and the number of trailing zeros of a product is the smaller of its total 2s and total 5s — worth saying when the product is not a factorial.

How the diagnostic asks it

One question from the Aptitude bank, exactly as a sitting would show it. The bank has 4 on number systems & divisibility and 60 across Aptitude.