the sum of the digits of a two-digit number is 10. If the given number is greater than its reverse number by 18. Find the given number
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Answer:
Step-by-step explanation:
Let a and b be the tens and ones digit respectively
So, the number is 10a + b
And the number obtained by reversing the digits is 10b + a
10a + b = 10b + a + 18
10a - a - 10b +b = 18
9a - 9b = 18
a - b = 2
a + b = 10
Adding eqn 1 and 2,
2a = 12
a = 6
6 + b = 10
b = 4
So, the number is 64
Hope for brainliest!!!
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