The Sum of digits integers X is 2 where 10^10 < X < 10^11 find the number of arrangements of X?
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We can rearrange the expression for x as 27∗(2∗5)1327∗(2∗5)13 so that (2∗5)13(2∗5)13 is equal to 1 followed by 13 zeros so that when it is multiplied by any number is equal to that number followed by 13 zeros. Then, 27=12827=128 so that multiplied by the power of 10 becomes 128 and 13 zeros. Since we have to sum the digit of this number, we get 1+2+8+13*0 = 11.
Answer 11
nandhinj:
tq
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