Each of the digits 7, 5, 8, 9 and 4 is used only once to form a three

Let a,b,c,d, and e represent a distinct digit from the above set.
abc
+de
555

Then 100*a+10*(b+d)+c+e=555
The terms 100*a and 10*(b+d) must end with zero, because they are multiples of 100 and 10 respectively. So it follows that c+e must end with a unit digit of 5. The smallest c+e can be is 9(5+4) and the biggest it can be is 17(8+9). The only number with a units digit of 5 between 9 and 17 is 15. From the given set, it follows that either c=7 or 8 and e=8 or 7.
Then 100*a+10*(b+d)=540 and the remaining digits that can be used are 5,9, and 4. If a=5 then b+d should be equal to 4, but this is not possible from the remaining digits to choose from. a must be equal to 4. Therefore b=5 or 9 and d=5 or 9. There are 4 numbers that satisfy the above equation. Correct answer is D.
Similar question to practice: https://www.artofproblemsolving.com/wiki ... Problem_15

Another method would be to start from the hundreds digit.

A three digit number and a two digit number add to give 555. So the three digit number must be greater than 455 but less than 500 since you do not have smaller digits to make one of the early 500 numbers such as 510. So the hundreds place of three digit number must have 4.
You are left with 5, 7, 8, 9
Next, the units digits must add to 15 and there is only one way of doing that: 7+8 = 15. The two digits, 7 and 8 can be arranged in either way - 4 __ 7 or 4 __ 8 so there are two ways to choose the units digit.
The leftover digits are 9 and 5 which need to go to tens places. Again, there are two ways of arranging them 4 9 __ or 4 5 __.
In all, there are 2*2 = 4 ways of making a 3 digit and a 2 digit number such that the sum is 555.

Answer (D)

1. Since the sum is 555 and ends with 5, the two numbers that form the unit digits in the 3 digit and 2 digit number must be 8 and 7.
2, We can see that first digit of the three digit number has to be 4.
3. We can easily list the numbers as 498 and 57, 458 and 97, 497 and 58, 457 and 98

Hi..
1) Hundreds digit cannot be anything other than 4.. 1 way
2) Units digit can be ONLY 8 and 7, one in 3-digit number and other in 2-digit.... 2-ways
3) Tens digit - the remaining two can also be any one in 3-digit and other in 2-digit..2 ways

ans 1*2*2=4
D

example - 138+27 = 137+28= 128+37=127+38

Hey Bunuel,

Can you post some similar sums of this type on this thread? Thanks

The posts by Bunuel are so crisp and methodical ! Thanks for being on the forum.

Check other Addition/Subtraction/Multiplication Tables problems from our Special Questions Directory

Asked: Each of the digits 7, 5, 8, 9 and 4 is used only one 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?

7 + 8 = 15
9 + 5 = 14
4

4 (9+5)(8+7) = 555
There are 2 * 2 = 4 choices
Numbers = {(498,57),(497,58),(458,97),(457,98)}

IMO D

From the question, we can deduce that
1) The Hundreds place will consist of 4 + 1(carried forward)
2) The Units place can ONLY have 8 and 7 as no other pair of digits can form the sum with units place of '5'
3) Thus leaving us with 9 and 5 as Tens place

Now, # ways to arrange Units place = 2!
and # ways to arrange Tens place = 2!
Since both are independent events,
thus total number of arrangements = 2! * 2! = 4

Only 2 numbers (7 and 8) whose sum add upto a 3 digits numer with a unit digit 5.
The 100th digit of the 3 digit number can only be 4, since we need the total sum as 555.

Now the digits in the 10th position is between 5 and 9.

In total we have 4 possible options (498,57) (497,58) (457,98) (458,97).

Ans D

Hi Bunuel,
Could the integers so formed be negative as well?
For these given numbers it wont be possible (let me now otherwise). But say in a question the digits were 5,8,9,3,4 : 589 and -34 could add up to 555.
Is this correct line of thought?

Looking forward to hearing from you. Thanks!

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