Divisibility Rules • Topic 2 of 2

Advanced Divisibility

CAT pushes divisibility two steps further. First, COMPOSITE divisors: to test a divisor n, split it into coprime factors and test each. For 12 use 3 and 4 (HCF 1), for 15 use 3 and 5, for 72 use 8 and 9. The trap is choosing factors that share a common factor — 12 = 6 × 2 fails, because passing the 2-test and the 6-test does not force divisibility by 12 (24 vs 12). Always pick factors whose HCF is 1, and their product covers the full divisor. Second, the REVERSE problem: a number has an unknown digit and you must find the value(s) that make it divisible by a given number. Set up the rule as an equation in the missing digit, then solve over 0–9. For divisibility by 9, the digit sum must reach the next multiple of 9; for 11, the alternating sum must hit 0 or ±11; for 8, only the last three digits matter, so the unknown often sits inside that block. When two digits are unknown, two rules (say 8 and 9, for 72) pin them down.

✅ Solved examples

1. Is 1,03,824 divisible by 12?
12 = 3 × 4 (coprime). Digit sum = 1+0+3+8+2+4 = 18, divisible by 3. Last two digits 24 = 4 × 6, divisible by 4. Both pass, so yes — divisible by 12.
2. Find the digit A so that 6,7,A,2 is divisible by 3 (number is 67A2).
Digit sum = 6+7+A+2 = 15 + A must be a multiple of 3. A = 0,3,6,9 give sums 15,18,21,24, all multiples of 3. So A ∈ {0, 3, 6, 9}.
3. For what digit B is 4,5,B,8 (i.e. 45B8) divisible by 8?
For 8, check the last three digits 5B8. Sweeping B = 0..9, the multiples of 8 are 528 = 8 × 66 (B = 2) and 568 = 8 × 71 (B = 6). So B ∈ {2, 6}.
4. The number 7,86,5xy is divisible by 72. Find x and y.
72 = 8 × 9 (coprime). For 9: 7+8+6+5+x+y = 26 + x + y must be a multiple of 9, so x + y = 1, 10 or 19. For 8: the last three digits 5xy must be ÷ 8. Combining the two, 5xy = 528 works: 528 ÷ 8 = 66, and digit total 7+8+6+5+2+8 = 36 is a multiple of 9. So x = 2, y = 8.

✏️ Practice — try these, take hints as needed

1. Is 5,67,180 divisible by 15?
15 = 3 × 5 (coprime).
Ends in 0, so ÷ 5 passes.
Digit sum = 27, ÷ 3 passes.
Yes
2. Find digit P so that 3,4,P,5,6 (34P56) is divisible by 9.
Digit sum = 18 + P.
Must be a multiple of 9.
18 + P = 18 or 27.
P = 0 or 9
3. Is 2,16,000 divisible by 72?
72 = 8 × 9.
Last three digits 000 ÷ 8 = 0.
Digit sum = 9, ÷ 9 passes.
Yes
4. For what digit Q is 9,Q,4,5,3 (9Q453) divisible by 11?
Alternating sum from the right: 3 −5 +4 −Q +9.
= 11 − Q.
Set 11 − Q to 0 or ±11.
Q = 0
5. Why is testing divisibility by 12 using the factors 6 and 2 unreliable?
HCF(6, 2) = 2, not 1.
They are not coprime.
Example: 18 passes 2 and 6 but 18 ÷ 12 is not an integer.
6 and 2 are not coprime, so passing both does not force divisibility by 12 (e.g. 18). Use 3 and 4.

📝 Topic test — 8 questions

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