Question 1058555: A certain number of two digits is 7 times the sum of its digits and increases by 27 when it is reversed.What is the number? Found 2 solutions by math_helper, Alan3354:Answer by math_helper(2461) (Show Source):
You can put this solution on YOUR website! Let the two digit number be x
x = 10a + b where a,b both in [0,9] (i.e. they are digits)
x = 7(a+b) (7 times the sum of its digits)
x = 10b+a+27 (increases by 27 when digits are reversed)
Three equations, three unknowns.
Set the first two equal to each other (they both have 'x' on the left hand side so they are equal), this will give us a in terms of b:
10a+b = 7(a+b)
10a+b = 7a+7b
3a = 6b
a = 2b (*)
Similarly, set the bottom two equations equal to each other:
7(a+b) = 10b+a+27
7a+7b = 10b+a+27
6a = 3b+27
Now (*) lets us plug in "2b" for "a" in this last equation:
6(2b) = 3b + 27
12b = 3b+27
9b = 27 ==> b=3 ==> a=6
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Ans: the number is 36.
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Check:
Reversing the digits is supposed to increase the number by 27:
36 + 27 = 63 (ok)
You can put this solution on YOUR website! A certain number of two digits is 7 times the sum of its digits and increases by 27 when it is reversed.What is the number?
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increases by 27 when it is reversed --> the digits differ by 3
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--> find a multiple of 7 whose digits differ by 3
03 NG
14 NG
25 NG
36 NG
47 NG
58 NG
69 NG
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No solution.
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If it decreases by 27, then 63 works.
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The other tutor says 36, but 36 is not 7*(3+6)
