Question
Soham took a 3 digit number and formed two other 3 digit numbers using the digits of the original number. He then added the three resulting numbers and divided their sum by 37 . He concluded that the result was exactly divisible by 37, no matter what number he chose. Is Soham's statement true or false?
Answer: Option A
:
A
Let the number be abc.
abc = 100a + 10b + c.
By rearranging the digits, let us say the two other numbers formed are cab and bca.
Expressing these two numbers in the expanded form,
cab = 100c + 10a + b
bca = 100b + 10c + a
abc + cab + bca
= (100a + 10b + c) + (100c + 10a + b) + (100b + 10c + a)
= 100a + 10a + a + 10b + b + 100b + c + 100c + 10c
= 111a + 111b + 111c
= 111(a + b + c)
= 37 × 3 × (a + b + c), since 111 is the product of 37 and 3.
Hence the sum is divisible by 37.
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:
A
Let the number be abc.
abc = 100a + 10b + c.
By rearranging the digits, let us say the two other numbers formed are cab and bca.
Expressing these two numbers in the expanded form,
cab = 100c + 10a + b
bca = 100b + 10c + a
abc + cab + bca
= (100a + 10b + c) + (100c + 10a + b) + (100b + 10c + a)
= 100a + 10a + a + 10b + b + 100b + c + 100c + 10c
= 111a + 111b + 111c
= 111(a + b + c)
= 37 × 3 × (a + b + c), since 111 is the product of 37 and 3.
Hence the sum is divisible by 37.
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