the tens digit of a 2 digit number is 3 more than the units digit. the sum of this number and the reverse is 143. find the number
Answers
Let N be the two-digit number in question. Let a be the first (tens) digit of N, and let b be the last (units) digit of N.
Then N = 10a+b, where a and b are non-negative integers less than or equal to 9.
From the information given in the question,
a = b+3.
And 10b+a = 10a+b-27.
So 10b+b+3 = 10(b+3)+b-27.
So 11b+3 = 10b+10×3+b-27.
So 11b+3 = 11b+30-27.
So 11b+3 = 11b+3.
This equation is an identity; it is true for any value of b.
To check this, let N' be the two-digit number obtained from reversing the digits of N.
Then N' = 10b+a.
If a = b+3, then:
N'-N = (10b+a)-(10a+b) = 10b+a-10a-b = 9b-9a = 9(b-a) = 9×[b-(b+3)] = 9×(b-b-3) =9×(-3) = -27.
So N'-N = -27.
So N' = N-27. ✓
So any pair of values for a and b, such that a and b are non-negative integers less than or equal to 9, and a = b+3, will satisfy the conditions of the problem.
The values of N which satisfy the requirements of the question are thus 41, 52, 63, 74, 85 and 96.
Check:
41–27 = 14. ✓
52–27 = 25. ✓
63–27 = 36. ✓
74–27 = 47. ✓
85–27 = 58. ✓
96–27 = 69. ✓
Khadijah21
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