Jerome swapped the digits of a two digit number?

Jerome swapped the digits of a two digit number. The new number he obtained was 30 less than the original. What was the original number?

I am pretty sure that there is no solution but would like some feedback. Thank you!

4 Answers

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  • TomV
    Lv 7
    7 years ago
    Favourite answer

    Original number = 10a+b

    New number = 10b+a

    10b+a = 10a+b - 30

    9b = 9a - 30

    b = a - 30/9 : Impossible since a and b are single digit integers, and 30/9 is not an integer

    No solution for a difference of 30. The difference would have to be a multiple of 9 for there to be a solution.

  • Mewtwo
    Lv 5
    7 years ago

    TomV's answer is excellent in its simplicity. Just note a and b are integers between 1 and 9, inclusive.

    Let's try a somewhat different approach though. Let us prove that no solution exists. We will need to prove two statements:

    (a) For any nonnegative integer z, 9 divides 10^z - 1.

    (b) Suppose that k is an integer. If m is an integer whose digits are determined by transposing two of those from k, then k - m is divisible by 9.

    Proof of (a). Clearly, for z = 0, the statement is true as 10^0 - 1 = 0 = 9(0). Suppose, then, for integers 1, 2, 3, ..., k, that 9 divides 10^k - 1. Then there is an integer s such that 9s = 10^k - 1 --> 10^k = 9s + 1. So,

    10^(k + 1) - 1 = (10^k)(10) - 1

    = (9s + 1)(10) - 1

    = 9(10s + 1),

    and 9 divides 10^(k + 1) - 1. So, by induction, 9 divides 10^z - 1 for every nonnegative integer z.

    Proof of (b): Let k = a_n 10^n + a_(n-1)10^(n-1) + ... +a_s 10^s + ... + a_t 10^t + ... + a_1 (10) + a_0 where the a_i's are integers between 0 and 9. Then, the integer m is of the form m = a_n 10^n + a_(n-1)10^(n-1) + ... +a_t 10^s + ... + a_s 10^t + ... + a_1 (10) + a_0. So,

    k - m = a_s 10^s + a_t 10^t - a_t 10^s - a_s 10^t

    = (a_s - a_t)(10^s - 10^t).

    Clearly, if s = t, then no transposition occurs and k - m = 0, which is clearly divisible by 9. So, suppose that s > t (we will omit the last case of s < t for it is very similar to this one). Then,

    k - m = (a_s - a_t)(10^s - 10^t)

    = (a_s - a_t)[10^t (10^(s - t)) - 10^t]

    = 10^t (a_s - a_t)[10^(s - t) - 1]

    But, s - t > 0 --> 9 divides 10^(s - t) - 1. So, 9 must divide the difference k - m. In the case s < t, we get the same result. Therefore 9 divides any such difference k - m (a difference between numbers whose digits are transposed).

    _____________________________

    To finish this problem, note that 9 does not divide 30. But, you claim that the difference between a number and one of its transposes is 30. So, 9 divides 30. This is obviously false (Since 30 = 9(3) + 3)! As a result, this problem is unsolvable.

  • 7 years ago

    Let x = tens digit

    Let y = ones digit

    Original Number = 10x + y

    Number with Digits Reversed = 10y + x

    10x + y - 30 = 10y + x

    9x - 9y = 30

    You have one equation and two unknowns, so you need more info to solve it.

  • Juan
    Lv 6
    7 years ago

    there is no solution

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