Solve the given problem
Answers
The sum of the digits 1 to 7 is 45 which is divisible by 9. As noted elsewhere, when you move a digit from the ones column to the tens column, you increase the sum by 10−1=9 times that digit so the sum will still be divisible by 9. 100 is not divisible by 9 so simply shifting digits will never give a sum of 100.
I am not sure if adding a decimal point still counts as just “using the numbers 1 to 9” but it is easy to see that shifting digits to the right of the decimal point will not work either since, by similar logic, the result will still be divisible by 7
I am guessing that using other operators like factorial, negative, brackets, and square root are not allowed. If you could use them, then it would be easy to create solutions like this:
1+2+43+(5∗6)+7+8+9=100
The other way of using the digits would be to use them as exponents. If we do this, then it is easy again to create a solution by first using the trick of shifting to the tens column to make a number that is larger than 100 but still very close and then reducing the sum by using digits to be exponents of 1. Here is a solution that follows the rules:
153+2+4+6+8+79=100
I got there by noting that summing 1 to 9 gives 45, so we want to add 55 to this total. Shifting the 7 to the tens column will add 63 for a total of 108. Then I moved two digits totaling 8 to the exponent of 1.
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the answer of thus questition is 1