What is the least multiple of 41 whose digits consist only of 1's

11111

Step-by-step explanation:

Multiple of a number is multiplication of the given number with any integer. These integers can be positives as well as negatives. Hence Multiples of any number can either be positive,negative or zero.

According to question,

We need to find a multiple of 41 which contains only 1's.

Hence,  we will start by dividing a number greater than 41 which only contains 1's. i.e 111 and keep on adding 1 at unit place until the number is divisible by 41.

111÷41 gives 29 as remainder, hence is not completely divisible by 41.

∴ we add 1 to its unit place and use the same process for 1111.

1111÷41 is also not divisible by 41 as it leaves 4 as remainder. Hence we add 1 to its unit place and check for 11111.

11111÷41=271. Hence, 11111 is completely divisible by 41.

∵ 11111 is the least number which contains only 1's and is a multiple of 41

nswer: a   y-axis

step-by-step explanation: