After actual multiplication we get 9 as ans. But, can this be done ,using some properties of the numbers? Rather than actual multiplication? Thanks in advance.
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You are asking for the digit sum of your number.It can be shown that the digit sum of any number is just the remainder when the number is divided by 9. Your number is clearly divisible by 9 because of the factor of 3^3 and so the remainder on division by 9 is 0,this is the same as a remainder of 9, so the digit sum is 9.
Another way is as follows:2^2*3^3*4^4*5^11 = 2^2*3^3*2^8*5^11 = 3^3*5*2^10*5^10
= 27*5*10^10 = 1350000000000.You can now see that the digit sum is 1 + 3 + 5 = 9.
Another way is as follows:2^2*3^3*4^4*5^11 = 2^2*3^3*2^8*5^11 = 3^3*5*2^10*5^10
= 27*5*10^10 = 1350000000000.You can now see that the digit sum is 1 + 3 + 5 = 9.
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2^2 = 4
3^3 = 9
4^4 = 16
5^11 = 48,828,125
48828154
4 is the result of adding the digits
of the above number.
3^3 = 9
4^4 = 16
5^11 = 48,828,125
48828154
4 is the result of adding the digits
of the above number.
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any number divisible by 2 or more factors of 3 has a total of 9