HOT Competition: Set A consists of five integers: 2, 4, 5, 6, and 7. X

Set A consists of five integers: 2, 4, 5, 6, and 7. X is a three digit number formed by using the digits from set A and Y is a two digit number formed by using the remaining digits from set A, such that all the digits are used only once. For example, X = 245 and Y = 67. If X + Y = 501, how many such pairs of integers can be formed?


For the sum to be 501, we need 4 in the hundred's place in 4xx and yy format.

Of the remaining digits only 5 & 6 add up to leave 1 in 1's place. That leaves us with 2 & 7 in the 10's place.

So, the possible combinations are (425, 76), (426, 75), (475, 26) & (476, 25).

Hence, the answer is C

If X+Y = 501
Either of X or Y have to an odd number. Other will be even no.
Plus the three digit number will never be 5xx,6xx,7xx because then the number itself will be more than sum.
So the hundred digit will be either 2 or 4
Let X = 4xx,2xx

The unit digit of both x and y have to be either 6 or 5
Because its sum give 1 as unit digit.
7&4 pain can't work as 4 is the hundred digit.

So X = 4x6 or 4x5
Y = y5 or y6

Put X and Y as rest no.(2&7) One by one
1) Let X = 475 and y will be 26
X+Y = 501
2) Let X = 425 and Y will be 76
X+Y = 501
3) Let X = 426 and Y will be 75
X+Y = 501
4) Let X = 476 and Y will be 25
X+Y = 501

There are 4 such pairs.
Answer must be C

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4 pairs can be formed : (425,76) (426,75) ( 475,26) (476,25).. IMO 4

Set A consists of five integers: 2, 4, 5, 6, and 7. X is a three digit number formed by using the digits from set A and Y is a two digit number formed by using the remaining digits from set A, such that all the digits are used only once. For example, X = 245 and Y = 67. If X + Y = 501, how many such pairs of integers can be formed?

A. None
B. One
C. Four
D. Six
E. Eight

Look For Sum to be 501

Hundredth Digit of X should be 4


Now for unit digit of sum is 1
It means
Unit digit of X and Y is 4 or 7 and 5 or 6 (Vice versa)


As we cannot use 4 as unit digit
we will use 5 or 6 as unit digits of X and Y
As we have 1 choice for hundredth digit of X =4
We have 2 possibility for tenth digit of X= 2 or 7
We have 2 possibility for unit digit of X= 5 or 6

We have 2 possibility for tenth digit of Y= 2 or 7
We have 2 possibility for unit digit of Y= 5 or 6


Only 4 Cases are possible
Possibilities are

425+76
426+75
475+26
476+25

You can not form any 2 digit no with 4. So options available : 2, 5, 6, 7

Now if you keep in units digit 2 then balance would be 9 and that is not possible
Further any number formed with 6 will need 3 as the difference would be in thirties and this is not possible.
Now same for numbers with 5, difference would be in fourties and 2 4's are not allowed

So numbers remaining : 70's and 20's

76 and 75 are possible
26 and 25 are possible

So 4

Set = {2,4,5,6,7}
Let us assume
X = abc
Y = ef

Since x+y= 501 (ones place = 1)
i.e. c+f = _1
Therefore c+f = 5+6 = 6+5
-------------------> 2 cases

501 (10th place = 0)

therefore b+e = 1(carry forward from 5+6)+7+2 = 1+2+7
-------------------> 2 cases
and a = 4
--------------------
Therefore total 4 cases are possible.

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2 units digit combo possible for this are 5,6 and 7,4

now for 5,6 one combination exists: 425+76
for 7,4 there is no combo as 4 has to be in units digit and there is no other combo can add up to 501

hence answer is one

Given digits are: {4, 5, 7, 8, 9}.

In order the units digit of the sum to be 5, the units digits of the two integers must be 7 and 8. Thus 2 cases are possible:

AB7
+C8
___
555

OR:

AB8
+C7
___
555

Clearly A must be 4:

4B7
+C8
___
555

OR:

4B8
+C7
___
555

B and C can be 5 and 9 or vise-versa.

Therefore there are total of 4 pairs: (457, 98), (497, 58), (458, 97), and (498, 57).

Answer: D.

this question can be solved using basic counting method.
For X+Y = 501
and X is a 3 digit number
and Y is a 2 digit number
all the digits are used only once
Digits can be formed using 2, 4, 5, 6, 7
Obviously, X cant be 5ab , 6ab, 7ab (all 3 are > 501) or 2ab (as y is just a 2 digit number)

Also, 501 unit digit is 1, the only 2 digit at unit place that after adding up will give 1 is 5 and 6

So, X = 4a5, 4a6
possible cases: 425, 426
475, 476

In that case, when x is above of 4 numbers, Y can be
501-425= 76 (possible)
501-426= 75 (possible)
501-475= 26 (possible)
501-476= 25 (possible)

Total 4 pair of integers are possible.
hence, Answer is C

The numbers we can use are: 2, 4, 5, 6, and 7.
We need to get 501 with the addition of two numbers consisting of above digits.

Let’s start with the three-digit number. It cannot be 2 because the addition of any two-digit number can result in at most 3XY, but not 501.
It cannot be 5, 6, and 7 either because in such case we get a number greater than 501.
So, only 4XY works.

Next, the sum of the last digits should give X1, and the only contenders are 5 and 6. We can write either 4X6 + Y5 or 4X5 + Y6. X and Y can be either 2 or 7. Hence, overall we have 4 cases:

426 + 75; 425 + 76; 475 + 26; 476 + 25

Hence C

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Set A consists of five integers: 2, 4, 5, 6, and 7. X is a three digit number formed by using the digits from set A and Y is a two digit number formed by using the remaining digits from set A, such that all the digits are used only once. For example, X = 245 and Y = 67. If X + Y = 501, how many such pairs of integers can be formed?

A. None
B. One
C. Four
D. Six
E. Eight

Given digits are: (2, 4, 5, 6, 7).

In order the units digit of the sum to be 1, the units digits of the two integers must be 5 and 6 OR 7 and 4.
Thus 2 cases are possible for 5 and 6. But we don't take into consideration 7 and 4 possibility, because in this case we cannot get sum 501 anyway.
So:

Consider last digit of X and Y as 6 and % ans vise versa:

AB5
+C6
____
501

AB6
+C5
____
501

A is most probably 4
so:

4B5
+C6
____

1) 475+26=501
2) 425+76=501

4B6
+C5
____
501

1) 476+25=501
2) 426+75=501

(C) is the answer.

A {2,4,5,6,7}

Set Y = {24, 25, 26, 27, 42, 45, 46, 47, 52, 54, 56, 57,62,64,65,67, 72, 74, 75, 76}

Set X{ Remaining three digits like 567, 467......}

X + Y = 501.

=> X + 24 = 501 => X = 477 (Not in Set X)

=> X + 25 = 501 => X = 476

=> X + 26 = 501 => X = 475

=> X + 27 = 501 => X = 474(Not in Set X)

=> X + 42 = 501 => X = 459 (Not in Set X)

=> X + 45 = 501 => X = 456 (Not in Set X)

=> X + 46 = 501 => X = 455(Not in Set X)

=> X + 47 = 501 => X = 454(Not in Set X)

=> X + 52 = 501 => X = 449(Not in Set X)

=> X + 54 = 501 => X = 447(Not in Set X)

=> X + 56 = 501 => X = 445(Not in Set X)

=> X + 57 = 501 => X = 454(Not in Set X)

=> X + 62 = 501 => X = 439(Not in Set X)

=> X + 64 = 501 => X = 437(Not in Set X)

=> X + 65 = 501 => X = 436(Not in Set X)

=> X + 67 = 501 => X = 434(Not in Set X)

=> X + 72 = 501 => X = 429(Not in Set X)

=> X + 74 = 501 => X = 427(Not in Set X)

=> X + 75 = 501 => X = 426

=> X + 76 = 501 => X = 425

Total 4 pairs of integers.

Answer C

The only problem is just stumbling on the first combinatioin initially i thought it was one however upon further inspectino i figured it out as (426, 75 ) , (475, 26), (476, 25) , (476 , 25) it took me around 3 minute to figure out the first the rest simply fell in place

5,6,7 cannot be the starting digits since ot will overshoot the sum of 501
2 cannot be used since then the value of 501 cannot be attained

It can start from 426, 75 a possibility that satisfies
remaining numbers can be formed by switching the unit digit place and tens place in 2*2 ways

=4
Therefore IMO C

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