Question

find the smallest number which, on being added 23 to it , is exactly divisible 32 and 90 ​

Answer 1

Answer:

First, we need to find the LCM of 24, 36 and 48. The LCM of 24, 36 and 48 is 144.

Now, we shall find the difference of 24 and 21 which is 3. Similarly, the difference of 36 and 33 and also that of 48 and 45 is 3 respectively.

The difference between the LCM and 3 is the smallest number.

So, 144 - 3 = 141.

Hence, 141 is the smallest number when divided by 24, 36 and 48 leave remainders 21, 33 and 45 respectively.

Answer 2

$\huge\color{indigo} \: Answer$

Step-by-step explanation:

First, we have to find the smallest number which is divisible by 32 and 90.

i.e. we have to find the LCM of 32 and 90.

which is -

2×2×2×2×2×3×3×5 = 1440

Now, subtract 23 from 1440

1440-23= 1417

☆ Hence, 1417 is the smallest number which, on being added 23 to it , is exactly divisible 32 and 90 .

☆ Rechecking,

1417 + 23 = 1440 is divisible by both 32 and 90.

hope it will help you