30 practice questions on divisibility-tests from the Number Play section of the UPSC Prelims syllabus. 30 come with a written explanation. Try the sample set below - the answer stays hidden until you ask for it.
What is the smallest three-digit number divisible by both $6$ and $11$?
Correct answer: A - $132$
A number divisible by both is a multiple of the LCM $66$, and the three-digit multiples of $66$ start at $132$. $110$ is a multiple of $11$ but is not divisible by $3$, $102$ is a multiple of $6$ but not of $11$, and $198$ works but is larger than $132$.
Assertion (A): To test whether a number is divisible by $4$, it is enough to look at its last two digits. Reason (R): $100$ is a multiple of $4$, so the hundreds, thousands and higher parts of any number are already multiples of $4$.
Correct answer: A - Both A and R are true, and R is the correct explanation of A
Since $100 = 4 \times 25$, everything above the tens place contributes a multiple of $4$, so only the last two digits can affect the remainder. That is precisely why the two-digit test works, so R explains A and both are true.
Assertion (A): Every number divisible by $3$ is divisible by $6$. Reason (R): A number divisible by $6$ must be divisible by both $2$ and $3$.
Correct answer: D - A is false, but R is true
A is false: $9$ is a multiple of $3$ but not of $6$, because it is odd. R is true, since $6 = 2 \times 3$ forces both factors. So the correct choice is the one that calls A false and R true.
Assertion (A): If the digit sum of a number is a multiple of $3$, the number is a multiple of $3$. Reason (R): If the digit sum of a number is a multiple of $2$, the number is a multiple of $2$.
Correct answer: C - A is true, but R is false
A is the standard test for $3$ and is true. R is false: the digits of $13$ add to $4$, a multiple of $2$, yet $13$ is odd. Divisibility by $2$ depends only on the last digit, so no digit-sum test works for it.
A number is divisible by $4$ and also by $6$. Which divisor is it certain to have?
Correct answer: D - $12$
The number must be a multiple of the LCM of $4$ and $6$, which is $12$. Multiplying instead gives $24$, but $12$ itself is divisible by $4$ and $6$ and not by $24$. The same example rules out $8$ and $18$.
Sohan tests $2\,348$ for divisibility by $8$ by adding its digits: $2+3+4+8 = 17$, which is not a multiple of $8$, so he says no. What is the correct verdict?
Correct answer: A - His answer happens to be right, but his method is wrong: the test for $8$ uses the last three digits, and $348$ is not a multiple of $8$
Dividing $348$ by $8$ leaves remainder $4$, so $2\,348$ is indeed not a multiple of $8$, but the digit sum is irrelevant here. The digit-sum test works only for $3$ and $9$, because $10$ leaves remainder $1$ on division by those, which is not the case for $8$.
A whole number is divisible by $4$ when the number formed by its
Correct answer: C - last two digits is divisible by $4$
Every hundred is a multiple of $4$, so only the last two digits can change the remainder. The last digit alone is not enough: $12$ and $14$ both end in an even digit, yet only $12$ is a multiple of $4$. The digit-sum test belongs to $3$ and $9$, and the leading digits play no part.
A four-digit number has digits $7$, $a$, $3$, $6$ in that order and is divisible by $9$. How many different digits can $a$ be?
Correct answer: A - $1$
The digit sum is $7 + a + 3 + 6 = 16 + a$, which must be a multiple of $9$. Since $a$ is a digit, $16 + a$ lies between $16$ and $25$, so the only multiple of $9$ available is $18$, giving $a = 2$. That is exactly one value, not two or three, and a solution does exist, so $0$ is wrong.
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