The tens digit of a two-digit positive integer is 2 more than three times the ones digit. If the digits are interchanged, the new number 13 less than half the given number. Find the given integer. (Hint: Let x = tens-place digit and y = ones-place digit, then 10x + y is the number.)
i think 82 is the answer.
unit digit: x
tens digit: y
the actual number: 10y + x
number reversed: 10x + y
"the new number 13 less than half the given number" ---> 10x+y < (1/2)(10y+x) by 13
so , take 13 away from the larger part to make them equal.
10x + y = (1/2)(10y+x) - 13
20x + 2y = 10y + x - 26
19x -8y = -26
also y = 3x + 2
19x - 8(3x+2) = -26
19x - 24x - 16 = -26
-5x = -10
x = 2 , then y = 3(2)+2 = 8
the original number is 82
check: number reversed = 28
half the given number = 41
is half the given number (41) less than the number reversed (28) by 13 ??
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