Question

If a 3 digit number ‘abc’ is divisible by 11, then _________
(a) a + b + c is a multiple of 11
(b) a + b – c is a multiple of 11
(c) a – b + c is a multiple of 11
(d) a – b – c is a multiple of 11

Answer 1

Hint: First of all, consider the number ‘abc’ given in the question. Now, number the digits starting from units place. Now, add the digits at even places and add the digits at odd places separately and take their difference which would be divisible by 11 according to its divisibility rule.
Complete step-by-step answer:
We are given a 3 digit number ‘abc’’ which is divisible by 11. Then we have to find the expression in which the digits a, b, and c must combine to be the multiple of 11. Before proceeding with the question, let us see what divisibility rule is.
A divisibility rule is the shorthand way of determining whether a given integer is divisible by a fixed divisor without performing the division, usually by examining its digits.
For example, for a number to be divisible by 2, it must have either 0, 2, 4, 6, and 8 as its last digit.
Now, let us see the divisibility rule of 11. Divisibility rule by 11 says that if the modulus of the difference of the sum of the digits at even places and digits at odd places of the given number is divisible by 11, then that number is divisible by 11 completely.
For example, we have a number 15081. Let us do the numbering of digits
$${\displaystyle \frac{1}{5}}{\displaystyle \frac{5}{4}}{\displaystyle \frac{0}{3}}{\displaystyle \frac{8}{2}}{\displaystyle \frac{1}{1}}$$
Here, the sum of the digits at even places = 8 + 5 = 13 and the sum of the digits at odd places = 1 + 0 +1 = 2.
Now, difference = 13 – 2 = 11
Here, we know that the difference that is 11 is divisible by 11. So, the number 15081 would also be completely divisible by 11 and so we get,
$${\displaystyle \frac{15081}{11}}=1371$$
Now, let us consider our question. We are given a number ‘abc’. Let us do the numbering of its digits.
$${\displaystyle \frac{a}{3}}{\displaystyle \frac{b}{2}}{\displaystyle \frac{c}{1}}$$
Here, the sum of the digits at odd places = (a + c).
Sum of digits at even places = b
Now, difference = (a + c) – b or b – (a + c) depending upon which is positive.
We are already given that ‘abc’ is divisible by 11. So, we can say that (a + c – b) or (b – a – c) would also be divisible by 11 or would be a multiple of 11.
Hence, the option (c) is the right answer.

Note: In this question, many students think abc as $$a\times b\times c$$ which is wrong because here a, b and c are digits of number abc. Also, students are advised to always start the numbering from the unit’s digit to correctly solve the question. Also, the difference taken should be positive. So, take the difference accordingly by subtracting a smaller number from the bigger number. In the above question, if we would have (– a + b – c) as an option, then that would also be correct because both the differences would be divisible by 11.