SHSAT Math Quiz: Divisibility Rules
3 questions · exam conditions
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Divisibility RulesQuestion 1 of 3

A number MM has the property that when any digit is removed from MM, the resulting number is divisible by 7. If M=abcd‾M = \overline{abcd} is a four-digit number, which of the following must be true?

Each of bcd‾\overline{bcd}, acd‾\overline{acd}, abd‾\overline{abd}, and abc‾\overline{abc} leaves the same remainder when divided by 7
MM itself must be divisible by 7
The sum a+b+c+da + b + c + d must be divisible by 7
At least three of the digits aa, bb, cc, dd must be the same
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SHSAT Math Quiz

SHSAT Math Quiz: Divisibility Rules

Practice Divisibility Rules in SHSAT Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Divisibility Rules, giving you a quick way to practice the rules, question types, and explanations that matter most for SHSAT Math.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A number MM has the property that when any digit is removed from MM, the resulting number is divisible by 7. If M=abcd‾M = \overline{abcd} is a four-digit number, which of the following must be true?

  1. Each of bcd‾\overline{bcd}, acd‾\overline{acd}, abd‾\overline{abd}, and abc‾\overline{abc} leaves the same remainder when divided by 7
  2. MM itself must be divisible by 7 (correct answer)
  3. The sum a+b+c+da + b + c + d must be divisible by 7
  4. At least three of the digits aa, bb, cc, dd must be the same
Explanation: Let M=abcd‾=1000a+100b+10c+dM = \overline{abcd} = 1000a + 100b + 10c + d. When we remove each digit, we get: bcd‾=100b+10c+d\overline{bcd} = 100b + 10c + d, acd‾=100a+10c+d\overline{acd} = 100a + 10c + d, abd‾=100a+10b+d\overline{abd} = 100a + 10b + d, abc‾=100a+10b+c\overline{abc} = 100a + 10b + c. All of these are divisible by 7. Now, 1000≡6(mod7)1000 \equiv 6 \pmod{7}, 100≡2(mod7)100 \equiv 2 \pmod{7}, 10≡3(mod7)10 \equiv 3 \pmod{7}. So M≡6a+2b+3c+d(mod7)M \equiv 6a + 2b + 3c + d \pmod{7}. The four three-digit numbers are: bcd‾≡2b+3c+d≡0(mod7)\overline{bcd} \equiv 2b + 3c + d \equiv 0 \pmod{7}, acd‾≡2a+3c+d≡0(mod7)\overline{acd} \equiv 2a + 3c + d \equiv 0 \pmod{7}, abd‾≡2a+2b+d≡0(mod7)\overline{abd} \equiv 2a + 2b + d \equiv 0 \pmod{7}, abc‾≡2a+2b+c≡0(mod7)\overline{abc} \equiv 2a + 2b + c \equiv 0 \pmod{7}. From the first two: 2b+3c+d≡2a+3c+d(mod7)2b + 3c + d \equiv 2a + 3c + d \pmod{7}, so 2b≡2a(mod7)2b \equiv 2a \pmod{7}, giving b≡a(mod7)b \equiv a \pmod{7}. From the first and fourth: 2b+3c+d≡02b + 3c + d \equiv 0 and 2a+2b+c≡02a + 2b + c \equiv 0, so 2b+3c+d≡2a+2b+c(mod7)2b + 3c + d \equiv 2a + 2b + c \pmod{7}, giving 2c≡2a(mod7)2c \equiv 2a \pmod{7}, so c≡a(mod7)c \equiv a \pmod{7}. From the first and third: 2b+3c+d≡02b + 3c + d \equiv 0 and 2a+2b+d≡02a + 2b + d \equiv 0, so 3c≡2a(mod7)3c \equiv 2a \pmod{7}. Since c≡a(mod7)c \equiv a \pmod{7}, we have 3a≡2a(mod7)3a \equiv 2a \pmod{7}, so a≡0(mod7)a \equiv 0 \pmod{7}. This means a≡b≡c≡0(mod7)a \equiv b \equiv c \equiv 0 \pmod{7}, and from 2b+3c+d≡02b + 3c + d \equiv 0, we get d≡0(mod7)d \equiv 0 \pmod{7}. Therefore, M≡6⋅0+2⋅0+3⋅0+0≡0(mod7)M \equiv 6 \cdot 0 + 2 \cdot 0 + 3 \cdot 0 + 0 \equiv 0 \pmod{7}. So MM must be divisible by 7.

Question 2

A positive integer nn satisfies the condition that n2n^2 is divisible by 72. What is the smallest possible value of nn?

  1. n=6n = 6
  2. n=12n = 12 (correct answer)
  3. n=18n = 18
  4. n=36n = 36
Explanation: We need n2n^2 to be divisible by 72. First, let's find the prime factorization of 72: 72=8×9=23×3272 = 8 \times 9 = 2^3 \times 3^2. For n2n^2 to be divisible by 23×322^3 \times 3^2, we need n2n^2 to contain at least 23×322^3 \times 3^2 as factors. If n=2a×3b×n = 2^a \times 3^b \times (other primes), then n2=22a×32b×n^2 = 2^{2a} \times 3^{2b} \times (other primes)2^2. For 232^3 to divide n2=22a×⋯n^2 = 2^{2a} \times \cdots, we need 2a≥32a \geq 3, so a≥1.5a \geq 1.5. Since aa must be an integer, a≥2a \geq 2. For 323^2 to divide n2=32b×⋯n^2 = 3^{2b} \times \cdots, we need 2b≥22b \geq 2, so b≥1b \geq 1. Therefore, the smallest nn has a=2a = 2 and b=1b = 1, giving n=22×31=4×3=12n = 2^2 \times 3^1 = 4 \times 3 = 12. Let's verify: n2=122=144=24×32n^2 = 12^2 = 144 = 2^4 \times 3^2. We need 144144 to be divisible by 72=23×3272 = 2^3 \times 3^2. Since 144=24×32144 = 2^4 \times 3^2 contains 23×322^3 \times 3^2 as a factor (because 4≥34 \geq 3 and 2≥22 \geq 2), 144144 is indeed divisible by 7272. Checking the other options: (A) n=6=2×3n = 6 = 2 \times 3, so n2=36=22×32n^2 = 36 = 2^2 \times 3^2. For divisibility by 72=23×3272 = 2^3 \times 3^2, we need at least 232^3, but 3636 only has 222^2. So 3636 is not divisible by 7272. (C) n=18=2×32n = 18 = 2 \times 3^2, so n2=324=22×34n^2 = 324 = 2^2 \times 3^4. This has 222^2, but we need 232^3, so 324324 is not divisible by 7272. (D) n=36=22×32n = 36 = 2^2 \times 3^2, so n2=1296=24×34n^2 = 1296 = 2^4 \times 3^4, which is divisible by 7272, but this is larger than n=12n = 12.

Question 3

In the sequence 100,101,102,103,…,999100, 101, 102, 103, \ldots, 999, how many numbers are divisible by both 6 and 10?

  1. There are 30 such numbers (correct answer)
  2. There are 31 such numbers
  3. There are 32 such numbers
  4. There are 33 such numbers
Explanation: A number divisible by both 6 and 10 must be divisible by their LCM. Since 6=2×36 = 2 \times 3 and 10=2×510 = 2 \times 5, we have LCM(6,10)=2×3×5=30\text{LCM}(6,10) = 2 \times 3 \times 5 = 30. So we need to count multiples of 30 in the range [100,999][100, 999]. The smallest multiple of 30 that is ≥100\geq 100 is found by: ⌈100/30⌉×30=⌈3.333...⌉×30=4×30=120\lceil 100/30 \rceil \times 30 = \lceil 3.333... \rceil \times 30 = 4 \times 30 = 120. The largest multiple of 30 that is ≤999\leq 999 is found by: ⌊999/30⌋×30=⌊33.3⌋×30=33×30=990\lfloor 999/30 \rfloor \times 30 = \lfloor 33.3 \rfloor \times 30 = 33 \times 30 = 990. The multiples of 30 in this range are: 30×4,30×5,30×6,…,30×3330 \times 4, 30 \times 5, 30 \times 6, \ldots, 30 \times 33. The number of terms is 33−4+1=3033 - 4 + 1 = 30. Let's verify: the multiples are 120,150,180,…,990120, 150, 180, \ldots, 990. This is an arithmetic sequence with first term a1=120a_1 = 120, common difference d=30d = 30, and last term an=990a_n = 990. Using an=a1+(n−1)da_n = a_1 + (n-1)d: 990=120+(n−1)×30990 = 120 + (n-1) \times 30, so 870=(n−1)×30870 = (n-1) \times 30, giving n−1=29n-1 = 29, so n=30n = 30. Therefore, there are 30 numbers divisible by both 6 and 10 in the given range.