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The sum of a two digit ten digit number and a single digit number is 7. If you add 45 to the two digit number, it will be the number formed by transposing the two digit number and the ten digit number. Then the two digit number is () A. 16B. 25C. 52D. 61
Let one digit be a and ten digit be B, then the two digit is (10b + a). From the meaning of the question, we get a + B = 710B + A + 45 = 10A + B, and the solution is a = 6B = 1. So the two digit is: 10 × 1 + 6 = 16
The sum of a two digit ten digit number and a single digit number is 7. If you add 45 to the two digit number, it will be the number formed by transposing the two digit number and the ten digit number. Then the two digit number is ()
A. 16B. 25C. 52D. 61
Let one digit be a and ten digit be B, then the two digit is (10b + a). From the meaning of the question, we get a + B = 710B + A + 45 = 10A + B, and the solution is a = 6B = 1. So the two digit is: 10 × 1 + 6 = 16
The sum of a two digit single digit and a ten digit number is 7. Transpose the two digits to form a new two digit number, which is 45 larger than the original number
TWT
Let the number in ten be X
10x+7-x+45=x+10(7-x)
9x+52=70-9x
18x=18
x=1
7-1=6
A: the number is 16
For a two digit number, the number on the one digit is three times that on the ten digit number. If the number on the tens of digits is exchanged with the number on the one digit, the two digit number obtained is 54 times larger than the original number. How to find the original two digit number?
Let the original ten digit number be x and the individual digit number be 3x. From the meaning of the question, we can get: (3x × 10 + x) - (10x + 3x) = 54. The solution is: x = 3, the original number is 3 × 10 + 9 = 39. A: the original two digit number is 39
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