Question

Find the greatest number that will divide 121, 33 and 55 leaving 1, 3 and 5 as remainder.

Answer 1

Hint: Subtracting the remainder from the given numbers and finding the highest common factor gives the greatest number.

Complete step-by-step answer:
Let us consider the number 121 first,

Here it was given that 121 when divided by the greatest number leaves the remainder as 1.

Similarly the number 33 when divided by the greatest number leaves the remainder as 3.

In the case of 55, when divided by the greatest number the remainder is 5.

Now Considering 121 again.

The greatest number divides 121 and leaves the remainder as 1, that means we have to subtract 1 from 121.

\[121-1=120\]

Now writing the factors for 120 we get,

\[120=2\times 2\times 2\times 3\times 5\]



The greatest number divides 33 and leaves the remainder as 3, that means we have to subtract 3 from 33.

\[33-3=30\]

Now writing the factors for 30 we get,

\[30=2\times 3\times 5\]

The greatest number divides 55 and leaves the remainder as 5, that means we have to subtract 5 from 55.

\[55-5=50\]

Now writing the factors for 50 we get,

\[50=2\times 5\times 5\]

To find the greatest number that divides all the 3 numbers, we have to find H.C.F (Highest common factor).

\[120=2\times 2\times 2\times 3\times 5\]

\[30=2\times 3\times 5\]

\[50=2\times 5\times 5\]

H.C.F of 120,30,50 is \[2\times 5\]= 10

Therefore the greatest number that divides 121, 33, 55 by leaving remainder 1,3,5 is 10.

Note: The student can make a mistake by putting the answer as 11 because all the given numbers are divided by 11 which gives a remainder 0.