P&C q 18

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Saumyojit

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A seven-digit number comprises of only 2's and 3's. How many of these are multiples of 12?

  1. 11
  2. 12
  3. 10
  4. 22


A no is divisible by 12 if it is divisible by 3 and 4 .

A no is divisible by 4 if no formed by last two digits is divisible by 4.

So, 32 is always present in tens and units place.

So, we need such combination of 2 and 3 's in the first five places so that we get sum when added with 5 results in a sum which is divisible by 3.

How to do that?
 
Saumyojit said:
A seven-digit number comprises of only 2's and 3's. How many of these are multiples of 12?

  1. 11
  2. 12
  3. 10
  4. 22


A no is divisible by 12 if it is divisible by 3 and 4 .

A no is divisible by 4 if no formed by last two digits is divisible by 4.

So, 32 is always present in tens and units place.

So, we need such combination of 2 and 3 's in the first five places so that we get sum when added with 5 results in a sum which is divisible by 3.

How to do that?
Click to expand...
What would the sum of digits be if the first 5 digits are all 2?

How many of the first five digits can be changed to 3?

In each case, how many ways are there to place those 3's?
 
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