Question

Each of the digits 7, 5, 8, 9, and 4 is used only once to form a three-digit integer and a two-digit integer. If the sum of the integers is 555, how many such pairs of integers can be formed?

Answer 1

Answer:

4 pairs

Explanation:

Consider the ones digit of the sum. We have to find by how many ways, numbers ending in 5 can be formed using the digits 7, 5, 8, 9 and 4.

  ⇒ If the three digit integer formed ends in 4, then the two digit integer shall end in 1 or vice versa, to get the sum of the integers end in 5. But no digit 1 is there. So 4 can't be...

  ⇒ If the three digit integer ends in 5, then the two digit integer shall end in 0 or vice versa, to get the ones digit of the sum of integers be 5. But there's no digit 0. Thus 5 can't be.

  ⇒ If the three digit integer ends in 9, then the two digit integer shall end in 6 or vice versa. But there's no digit 6. Thus 9 can't be.

  ⇒ So the only possibility is that, the ones digit of the three digit integer and the two digit integer shall be 7 and 8 respectively, or vice versa.

Thus, 2 numbers at ones place.

Consider the tens digit of the sum. As the only possibility for the ones digit is being 7 or 8, thus 7 + 8 = 15, thus 1 of 15 is moved towards the tens digit and added there. Such that the possible digits at the tens place of the three digit integer and two digit integer must have sum ending in 4. So we have to find by how many ways, numbers ending in 4 can be formed using digits 5, 9 and 4. (7 and 8 are used at ones digit, and no digit repetition occurs.)

  ⇒ If the three digit integer has 4 at its tens place, then the tens digit of the two digit integer shall be 0, or vice versa. But there's no 0. So 4 can't be.

  ⇒ Thus the only possibility is for the tens digit of the three digit integer and the two digit integer being 5 and 9 respectively, and vice versa.

Thus, 2 numbers at tens place too.

Here we find that there is only possibility for the hundreds place of the three digit integer, and that is for being 4. Thus, 1 number at hundreds place of the three digit integer.

No. of total pairs = 1 × 2 × 2 = 4

   

And those are given below:

(457, 98)

(458, 97)

(497, 58)

(498, 57)