if the first and the last digit of a three-digit number differ by 6. find the difference between the number and the obtained by reversing it digit.
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let the first digit be a
.•.last digit = a-6
let middle number be b
real no. = 100a+10b+(a-6)
reversed no.=100(a-6)+10b+a =100a-600+10b+a
difference = (100a+10b+a-6)-(100a-600+10b+a)
= 100a+10b+a-6-100a+600-10b-a
= 594
the answer is 594...
I hope it helped you...
please mark it as brainliest...
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.•.last digit = a-6
let middle number be b
real no. = 100a+10b+(a-6)
reversed no.=100(a-6)+10b+a =100a-600+10b+a
difference = (100a+10b+a-6)-(100a-600+10b+a)
= 100a+10b+a-6-100a+600-10b-a
= 594
the answer is 594...
I hope it helped you...
please mark it as brainliest...
^_^
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