Classification of numbers, divisibility rules and factorials · PEA515

Natural numbers are 1, 2, 3 ...; whole numbers add 0; integers add the negatives. Rational numbers are p/q with q not 0. A prime has exactly two factors (1 and itself), so 1 is neither prime nor composite and 2 is the only even prime. Two numbers are co-prime when their HCF is 1.

Divisor	Test
2, 5, 10	last digit even, 0 or 5, 0
3, 9	digit sum divisible by 3, 9
4, 8	last two, three digits divisible by 4, 8
6	divisible by both 2 and 3
11	(sum of odd-place digits) minus (sum of even-place digits) is 0 or a multiple of 11

Factorial: n! = n x (n-1) x ... x 1, and 0! = 1. Trailing zeros of n! = floor(n/5) + floor(n/25) + floor(n/125) + ...; for 100! that is 20 + 4 = 24.

Exam trap: checking divisibility by 6 with only 2 and 3 is right, but checking by 12 with only 2 and 6 is wrong; use co-prime pairs such as 3 and 4.

Try two questions

Question 1 · Tier 4, Analyse

How many integers from 1 to 722 are divisible by neither 10 nor 6?

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A. Multiples of 10: 72, of 6: 120, of LCM 30: 24. Union = 168, so neither = 722 - 168 = 554.

Why not B: Subtracts both counts but forgets numbers divisible by both were removed twice.

Why not C: Gives the count divisible by at least one instead of neither.

Why not D: Removes only multiples of the LCM.

Question 2 · Tier 3, Apply

Find the smallest digit that replaces * so that the number 2635* is divisible by 3?

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C. Apply the divisibility rule of 3 to the visible digits and choose the smallest completing digit, which is 2.

Why not A: Picks the largest digit that works instead of the smallest.

Why not B: Gives the remainder of the other digits instead of the digit that completes the multiple.

Why not D: Arithmetic slip in the final step.

Key points

Try two questions