document.write( "Question 268409: If 10^51 – 9 is written as an integer in standard form , what is the sum of the integer’s digits
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Algebra.Com's Answer #196700 by jsmallt9(3759) $\"\"$ $\"About$

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$\"10%5E51\"$ is a 1 followed by 51 zeros. When we subtract 9 from it we will have to borrow. If you think about how borrowing works, you will realize that the 1 at the start will become a zero and that all the zeros except the last one (IOW 50 of them) will become 9's and the last one will become a 10. After we subtract the 9 we will get a 1 at the end with 50 9's in front of it. The sum of 50 9's and a 1 would be 50*9 + 1 = 450 + 1 = 451. \n" ); document.write( "

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