SOLUTION: A two digit number is six times the sum of its digits. If the digit in the units’
place is increased by 2 and the digit in the ten’s place is decreased by 2,
then the number so
Algebra.Com
Question 759870: A two digit number is six times the sum of its digits. If the digit in the units’
place is increased by 2 and the digit in the ten’s place is decreased by 2,
then the number so formed is 4 times the sum of its digits. Find the
original number.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A two digit number is six times the sum of its digits. If the digit in the units’place is increased by 2 and the digit in the ten’s place is
decreased by 2,then the number so formed is 4 times the sum of its digits.
Find the original number.
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Let the original number be 10t+u
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Equations:
10t+u = 6(t+u)
10(t-2)+(u+2) = 4(t-2+u+2)
----
Rearrange:
4t - 5u = 0
6t - 3u = 18
------
Modify for elimination:
4t - 5u = 0
4t - 2u = 12
-----
3u = 12
u = 4
----
Solve for "t":
4t = 5u
t = 5
----
Original Number: 10t+u = 54
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Cheers,
Stan H.
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