GMAT Club World Cup 2022 (DAY 10): Is the sum of four consecutive

Answer is B

(1) n(n+1)(n+2)(n+3) = 120k --> (n+3)! = 120 k
Insufficient

(2) XYZ = 120 k

4 * 5 * 6 * 7 = 840 = 120 * 7
5 * 6 * 7 * 8 = 1680 = 120 * 14
3 * 4 * 5 * 6 = 360 = 120 * 3
2 * 3 * 4 * 5 = 120 = 120 * 1
Their sums are not a multiple of 10

Thus, (2) alone is sufficient.

Asked: Is the sum of four consecutive integers a multiple of 10 ?

Let the consecutive integers be x, x+1, x+2 & x+3.
Is (x+x+1+x+2+x+3) = 4x + 6 is a multiple of 10 ?

(1) The product of the four integers is a multiple of 120.
x(x+1)(x+2)(x+3) = (x^2 + 3x)(x^2 + 3x+2) = (t-1)(t+1) = t^2 - 1; where t = x^2 + 3x +1
t^2 - 1 = 120k; t^2 - 120k - 1 = 0; \(t = \sqrt{120k+1}\) or \(t = -\sqrt{120k+1}\)
If k =1; t^2 = 121; For other values of k, t is NOT an integer. Therefore, x is NOT an integer.
t=x^2 + 3x +1 = 11; x^2 + 3x -10 = 0; (x+5)(x-2) = 0; x = 2 or x = -5
t=x^2 + 3x +1 = -11; x^2 + 3x +12 = 0; x is NOT an integer
Consecutive integers = {-5,-4,-3,-2} or {2,3,4,5}
Sum of 4 integers = (2+3+4+5) = 14 or -(2+3+4+5) = -14;
Sum of 4 integers is NOT a multiple of 10.
SUFFICIENT

(2) The product of the units digits of the four integers is a multiple of 120.
2*3*4*5 = 120 = 120*1 : Sum of unit digits = 2+3+4+5 = 15: NOT a multiple of 10
3*4*5*6 = 360 =120*3: Sum of unit digits = 3+4+5+6 = 18: NOT a multiple of 10
4*5*6*7 = 840 = 120*7: Sum of unit digits =4+5+6+7 = 22: NOT a multiple of 10
5*6*7*8 = 1680 = 120*14: Sum of unit digits = 5+6+7+8 = 26: NOT a multiple of 10
6*7*8*9 is NOT a multiple of 120
7*8*9*0 = 0 = 120*0: Sum of unit digits =7+8+9+0 = 24: NOT a multiple of 10
SUFFICIENT

IMO D

Is the sum of four consecutive integers a multiple of 10 ?

the sum must have 0 in the units digit for it to be a multiple of 10

(1) The product of the four integers is a multiple of 120.
if numbers are 1,2,3,4 sum = 10 and product = 24 hence Sufficient
but if numbers are 2,3,4,5 than sum = 14 and product = 120 Not sufficient
Therefore st 1 is not sufficient
(2) The product of the units digits of the four integers is a multiple of 120.
Clearly sufficient
Answer B

Using statement 1,

Some of the possible sets of values are:
{2, 3, 4, 5}, {3, 4, 5, 6}, {4, 5, 6, 7}, {5, 6, 7, 8}, {7, 8, 9, 10}, {8, 9, 10, 11}, {9, 10, 11, 12}, {10, 11, 12, 13}, and so on...

None of these add up to be a multiple of 10, and there's a reason why.
120 needs 3 twos, 1 three and 1 five. To be a multiple of 120 and be a consecutive number AND add up to be a multiple of 10 is not possible.

Hence, statement 1 is sufficient.

Statement 2 limits our possible sets to {2, 3, 4, 5}, {3, 4, 5, 6}, {4, 5, 6, 7}, and {5, 6, 7, 8}.

As we know, this isn't going to yield a sum which will be a multiple of 10.

Hence, statement 2 is sufficient as well.

The answer is option D.

Statement 1

\(120 = 5 * 3 * 2^3\)

This indicates that out of the four consecutive numbers, one number should be a multiple of 5.

The number 5 can come in any position, i.e. the series can either start with 5 or end with 5 or 5 can occupy any of the two middle positions.

Ex.

5 6 7 8
4 5 6 7
3 4 5 6
2 3 4 5

Alternatively, we can also have 10

10 11 12 13
09 10 11 12
08 09 10 11
07 08 09 10

To ascertain "Is the sum of four consecutive integers a multiple of 10 ?", we only care about the last digit. Hence in both the sets we see that the sum of the last digits is not a multiple of 10.

Hence this statement is enough to ascertain that the sum is not a multiple of 120

Statement 2

The analysis to this is similar to what we have already done. Infact, its a subset -

The number can have the unit digit in any one of the below series -

5 6 7 8
4 5 6 7
3 4 5 6
2 3 4 5

As discussed in statement 1 , the sum of the unit digits does not add to be a multiple of 10, hence this statement is also sufficient

IMO D

Correct answer : Choice D

Let x be the starting integer:
x, x+1 , x +2 ,x +3 will be the 4 integers

to find if x + x+1 +x + 2 + x+3 = multiple of 10
4x + 6 = 10k
with this equation, we can get to know which sum of 4 consecutive integers are multiples of 10 and which sets are not

1,2,3,4 = yes
2,3,4,5 = no
3,4,5,6 = no
4,5,6,7 = no
5,6,7,8 = no
6,7,8,9 = yes
.
.
.

now statement 1 says : The product of the four integers is a multiple of 120.
This can happen only on the "no" sets of numbers above. since sum of those numbers is not a multiple of 10 , statement 1 is sufficient

statement 2 says : The product of the units digits of the four integers is a multiple of 120.
The same logic can be applied here as well. hence statement 2 is sufficent too.

Hence correct answer is choice D

Is the sum of four consecutive integers a multiple of 10 ?

(1) The product of the four integers is a multiple of 120.
(2) The product of the units digits of the four integers is a multiple of 120.

statement 1 - abcd = 120x; have no idea about a,b,c,d & x

not sufficient

statement 2 - we have some idea about unit digits, unit digits should have 2, 3 and 5

we can calculate the sum of unit digits, if it has 0 we can say sum of the consecutive integer divisible by 10

sufficient

answer B

Answer D for me.

Explanations in the picture.



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Is the sum of four consecutive integers a multiple of 10 ?

(1) The product of the four integers is a multiple of 120.
(2) The product of the units digits of the four integers is a multiple of 120.

For the sum of 4 consecutive integers to be multiple of 10,the sum of the unit digits of the consecutive integers should be multiple of 10 as well
Now if we just consider the unit's digit for the 4 consecutive integers possible combinations would be
(1,2,3,4) (2,3,4,5) (3,4,5,6) (4,5,6,7) (5,6,7,8) (6,7,8,9)
Amongst these 6 for (1,2,3,4) and (6,7,8,9) - the sum is a multiple of 10
1) As per statement 1 the product of 4 integers is a multiple of 120
Now the product to be multiple of 120 the product of the digits of the unit's place must also be a multiple of 10
From the combinations given above for (2,3,4,5) (3,4,5,6) (4,5,6,7) (5,6,7,8) the product is multiple of 10.
and for these cases the sum is not a multiple of 10.
Thus the sum of 4 consecutive integers cannot be multiple of 10
Statement 1 is sufficient.
2)As explained above for the following cases :- (2,3,4,5) (3,4,5,6) (4,5,6,7) (5,6,7,8) product is multiple of 120.
But for none of the group the sum is multiple of 10.
thus the sum of four consecutive integers cannot be multiple of 10.
Statement 2 is sufficient.
D is the answer.

Let the four consecutive integers be n-1, n, n+1, n+2
Question Is sum of four consecutive integers a multiple of 10 ?
Or Is 4n+2 = 10K ? [where K is an integer]

Statement - 1
"The product of the four integers is a multiple of 120."

First note that in order for sum to be a multiple of 10, the units digit of the resulting sum of the 4 consecutive integers must be a 0. In other words the sum of the units digits of the 4 consecutive integer must end in a 0.

Second, if the product of 4 consecutive integers is to be a multiple of 120, then the 4 numbers must have 2,3,5 as distinct factors (120=2^3*3*5)

Now let's list down all the possibilities of units digit of the 4 consecutive integers. We are considering only positive values as negative values will have same results as positive values in all these cases.

0,1,2,3 [e.g. 10,11,12,13] -> This could lead to a multiple of 120 because factors 2,3 and 5 are present. But sum of the unit digits = 0 + 1+ 2+ 3=6, and thus the sum of the 4 consecutive integers will not lead to a multiple of 10 in this case.
1,2,3,4 -> Clearly, this can NOT be a multiple of 120, since there cannot be a 5 as a factor of these numbers [a number which has 5 as a factor must either end in 5 or end in 0]
2,3,4,5 -> This could lead to a multiple of 120 because factors 2,3 and 5 are present. But sum of the unit digits =Unit(2+3+4+5)=4, and thus the sum of the 4 consecutive integers will not lead to a multiple of 10 in this case.
3,4,5,6 -> This could lead to a multiple of 120 because factors 2,3 and 5 are present. But sum of the unit digits =Unit(3+4+5+6)=8, and thus the sum of the 4 consecutive integers will not lead to a multiple of 10 in this case.
4,5,6,7 -> This could lead to a multiple of 120 because factors 2,3 and 5 are present. But sum of the unit digits =Unit(4+5+6+7)=2, and thus the sum of the 4 consecutive integers will not lead to a multiple of 10 in this case.
5,6,7,8 -> This could lead to a multiple of 120 because factors 2,3 and 5 are present. But sum of the unit digits =Unit(5+6+7+8)=6, and thus the sum of the 4 consecutive integers will not lead to a multiple of 10 in this case.
6,7,8,9 -> Clearly, this can NOT be a multiple of 120, since there cannot be a 5 as a factor of these numbers [a number which has 5 as a factor must either end in 5 or end in 0]
7,8,9,0 -> This could lead to a multiple of 120 because factors 2,3 and 5 are present. But sum of the unit digits =Unit(7+8+9+0)=4, and thus the sum of the 4 consecutive integers will not lead to a multiple of 10 in this case.

We see in all cases the sum of the 4 consecutive integers cannot lead to a multiple of 10 if their product is a multiple of 120.
Hence SUFFICIENT.

Statement - 2
"The product of the units digits of the four integers is a multiple of 120."

What this statement says is basically an extension of what statement 1 said. In this case rather than considering 4 consecutive numbers like before we consider each case with their units digits and the same result as above follows.
Hence SUFFICIENT.

IMO Option D is the answer.

The only integers set forms that will support statement I are integers that have unit digits of the form below:

0, 1, 2,3
_1,0,1,2
-2,-1,0,1
-3,-2,-1,0
5,6,7,8
4,5,6,7
3,4,5,6
2,3,4,5

A number with the unit digit ending in 0 must be present to be a multiple of 120. A number with unit digit 5 or 0 must be part of the four numbers to be multiplied to give a multiple of 120.
The sum of any four consecutive integers including 5 or 0 will never be multiple of 10 .

Therefore statement I is sufficient.

Using the same analogy in Statement I, Statement II is sufficient.

For example 4+5+6+7 = 22
44+45+46+47 = 182

Both the sum above in bold will never be a multiple of 10.

Therefore, option D is the answer IMO.

1) Sufficient
2) Sufficient independent of (1)

Ans - D

Solution attached

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Let the four consecutive integers be n, n+1, n+2, n+3. So the sum = 4n + 6.
If the sum is a multiple of 10 then,
4n + 6 = 10k, where k is an integer.
So, 4n + 6 can have values as {10,20,30,40,50,60,...etc}
When we try to solve this equation we see that values of n for which n is an integer are:
n = {1, 6, 11, 16, 21, 26, 31,...}
So the consecutive integers could be:
1,2,3,4 or 6,7,8,9 or 11,12,13,14 or 16,17,18,19 etc.

From st. 1 we have the product of the four integers is a multiple of 120.
So, n*(n+1)*(n+2)*(n+3) = 120*k, k is a integer
120k = 2^3 * 3 * 5 * k

So we see that if the product is a multiple of 120, then one of the terms must have 5 in it.
But from the question stem we have identified that the four consecutive integers should not contain a 5 if their sum is a multiple of 10.
Hence from st. 1 we can conclude that if the product of the four integers is a multiple of 120, then their sum can not be a multiple of 10.
Hence this statement is sufficient.

From st. 2 we have the product of the units digits of the four integers is a multiple of 120.
In this case also we see that if the product of units digits is a multiple of 120 then also we would need atleast one of the terms to be ending with 5.
But from the question stem we have identified that the four consecutive integers should not contain a 5, if their sum is a multiple of 10.
So from this statement also we can say that their sum will not be a multiple of 10.
Thus this statement is sufficient.

Hence answer choice (D)

Imo D
Statement 1: The product of the four integers is a multiple of 120.
a * b**c*d = 120 k = (2^3)*3*5
There are only 2 set of consecutive numbers that satisfy above equation -
2,3,4,5 - Sum is 14
and 3,4,5,6 - Sum is 18
Both are not a multiple of 10.
Sufficient

Statement 2: The product of the units digits of the four integers is a multiple of 120.
Take one set of values- 120 =2*3*4*5 or 3,4,5,6 [ unit digits should also be consecutive]
We know the sum of these numbers are not a multiple of 10.
Lets take 52*53*54*55 (any random consecutive numbers)
The sum will be 50(4)+(2+3+4+5)
Clearly the sum is not divisible by 10
Sufficient

IMO D

Is the sum of four consecutive integers a multiple of 10 ?

Statement 1

(1) The product of the four integers is a multiple of 120.

120=2*2*2*3*5

Which means one of the four integers must be a multiple of 5 or 10

In both such cases the consecutive numbers will be ending with 2 3 4 5 6 7 8. Or 7 8 9 0 1 2 3
We can club any 4 of them to identify our numbers.
Eg 2345, 3456, 7890, 0123
In any case the sum will not give a 0 in its units place.
Hence never will the sum be a multiple of 10
Sufficient

Statement 2

(2) The product of the units digits of the four integers is a multiple of 120.

120= 2*3*4*5
This is the only consecutive units digits that will give a product of 120
Their addition also doesn’t yield a multiple of 10.

So never, thus Sufficient

Hence answer is D

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Answer: D

Is the sum of four consecutive integers a multiple of 10 ?

Let the four consecutive integers be x, x+1, x+2 and x+3
So, the sum = 4x + 6

We need to know if 4x + 6 = 10k ... where k can be any integer.

Now, 4x+6 = 10k => 2x + 3 = 5k
=> x = (5k - 3) / 2 .... where x should be integer
=> 5k - 3 should be even.
=> k needs to be odd.
When k = 1, x = 1
k=3, x=6
k=5, x=11
and so on..
We can observe that x increments with 5 for every odd k.

So, possible sets of consecutive integers where their sum = multiple of 10 will be
{1, 2, 3, 4}; {6,7,8,9}; {11,12,13,14}; {16,17,18,19} and so on..
Another way to interpret this is we need to find the set of numbers which fall between consecutive multiple of 5 and 10.
If the set of numbers contain any number which is multiple of 5 or 10 then that set will not have sum as multiple of 10.

From the given statements, we need to find if we can get above set of integers or not..i.e. whether the set of integers has atleast one of the number as multiple of 5 or 10.

(1) The product of the four integers is a multiple of 120.
=> x*(x+1)*(x+2)*(x+3) = 120n ....where n can be any integer.
It's clear that 120n is a multiple of 5 and 10.
Hence, at least one of the number will be a multiple of 5.
E.g. when n = 1, x = 1 and numbers are 2, 3, 4 and 5.
Since there will be at least one number which will be a multiple of 5 or 10, the required set of numbers will never have sum as a multiple of 10.
Sufficient.

(2) The product of the units digits of the four integers is a multiple of 120.
E.g. 2*3*4*5 = 120
For 12, 13, 14 and 15 --> 2*3*4*5 = 120
Again, its clear that at least one of the number of this set will be a multiple of 5 thereby implying that required set of number will never have sum as a multiple of 10.
Sufficient.