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# This isn't a riddle. It's my 8th grade math.

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In a two-digit number, the digit in the tens place value is 3 times the digit in the units place value. When the number is decreased by 54, the digits are reversed. What is the number?

Dec 1, 2015

#2
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Let the unit digit = x

Then, the tens digit  = 3x

And we have the following equation

[3x(10)  + x]   - 54  =  [ x(10) + 3x ]  simplify

30x + x   -   54   =  10x + 3x

31x - 54  = 13 x

18x  - 54  = 0

18x  = 54      divide both sides by 18

x  = 3         and 3x  = 3(3)   = 9

So

The larger number  is  3(3)(10) + 3   = 93

And the smaller number  is  10(3) + 3(3)  =  39

And their difference = 54

Just as Anonymous4338 found  !!!!!!   Dec 1, 2015

#1
+5

9 is 3 times more than 3, and when you subtract 54 from 93, you get 39.

Dec 1, 2015
#2
+5

Let the unit digit = x

Then, the tens digit  = 3x

And we have the following equation

[3x(10)  + x]   - 54  =  [ x(10) + 3x ]  simplify

30x + x   -   54   =  10x + 3x

31x - 54  = 13 x

18x  - 54  = 0

18x  = 54      divide both sides by 18

x  = 3         and 3x  = 3(3)   = 9

So

The larger number  is  3(3)(10) + 3   = 93

And the smaller number  is  10(3) + 3(3)  =  39

And their difference = 54

Just as Anonymous4338 found  !!!!!!   CPhill Dec 1, 2015
#3
+5

In a two-digit number, the digit in the tens place value is 3 times the digit in the units place value. When the number is decreased by 54, the digits are reversed. What is the number?

Let the first digit =n, then we have the 2nd digit=3n. But we have:

(3n*10) + n - 54=10n + 3n

30n + n - 54=13n

31n - 13n=54

18n=54

n=54/18

n=3 the first number

3 X 3=9 the second number, therefore the number is:

=93, because 93 - 54=39

Dec 1, 2015