IMAT 2015 Q9 [Digit Referance]

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When I made a hotel reservation online yesterday I was given an 8­digit booking reference which contained no zeros. It did, however, consist of three 2­digit odd numbers followed by the sum of these three numbers, and all eight digits were different.

The first digit of the booking reference was 4. What was the last digit?

A. 3
B. 5
C. 9
D. 7
E. 1

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Simple steps to solve word problems:

  • Underline key information
  • Determine what they are trying to ask, and what you will need to solve it
    • Eliminate any non-essential information
  • Draw a picture, graph, or equation
    • In moments of high stress like exam taking, always work with the paper they give you to avoid careless mistakes.
  • Solve.

Givens:

8 digit confirmation

All 8 digits are different

Format: 3 two-digit odd numbers, followed by their sum

First digit is 4

So: 4X + XX + XX = XX

First off, we know that the two-digit numbers are odd, so 1,3,5,7,9. This limits what we put as the first number of the double-digit numbers because we cannot repeat any numbers, and there are 3/5 of the odd numbers that will occupy the one’s column. We also need to be careful not to overshoot the total. Let’s take the most straightforward approach, a smart way of guessing. How can we guess smartly? We will try to make the smallest number possible. We will do this by putting the smaller numbers in as the tens unit (X_) and the larger numbers as the ones (_X). The numbers in the one’s column do not matter as much because we are limited in options, it has to be one of the 5 choices.

After trialing some numbers, you will realise that certain combinations are not allowed. For example, you cannot have 7 and 3 together in the first three one’s columns of the double-digit numbers because it results in the final answer repeating the third double-digit number’s one column.

Ex. X7 + X3 + XY = XY (Because 7 + 3 + Y = 10 + Y, resulting in Y being in the final answer).

After trialling numbers and realising what to avoid, you should end up with:

49 + 23 + 15 = 87

\fcolorbox{red}{grey!30}{Therefore the answer is D, because the final number is $7$.}
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3

why not
43 + 17 + 25 = 85

it’s the same method but it gives yet smaller value for the last 2 digit number.

4

hi @Karam
the question tells us that all 8 numbers have to be different
in your answer, the number 5 appears twice in 25 and in 85
so this cannot be a valid solution

5

ohhh I see, thanks!!

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