the sum of the digit of a two digit number is 6 on reversing its digits the new number is 18 less than the original number find the number
Answers
The sum of the digits is 6
Let the digit at ten's place be (6-x)
Let the value of the unit digit be x*1=x
let the value of the ten's digit be (6-x)*10
=60-10
original number= 60-10x+x =60-9x
on reversing the digits = 6-x*1 + x*10
6-x+10x
6+9x
ATQ
(60-9x)-18 = 6+9x
60-18-6 = 9x+9x
36 = 18x
x = 36/18
x = 2
So, therefore the original no is
(6-x)*10 + (x)*1
(6-2)*10 + (2)*1
4*10 + 2
40 + 2
42
Reversed no =24
I HOPE THIS HELPS
Given : The sum of the digit of a 2- digit number is 6. On reversing its digits, the number is 18 less than the original number
To Find : the number
Solution:
Let say original number = AB
A + B = 6
Reversed number = BA
On reversing its digits, the number is 18 less than the original number
10B + A = 10A + B - 18
=> 9(A - B) = 18
=> A - B = 2
2A = 8
=> A = 4
B = 2
Number = 42
Verification :
4 + 2 = 6
24 = 42 - 18
42 is the number whose sum of the digit is 6 and on reversing its digits, the number is 18 less than the original number
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