The sum of three integers is 40. The largest integer is 3

I did the same way.
Agree with B...
IMO:B

Let us assume 3 different integers to be x, y and z, in ascending order.

Given that x + y + z = 40;
We are give that z = 3y & x = z - 23 (or) x = 3y - 23;(just to keep a single variable for calculation)

3y - 23 + y + 3y = 40;
7y = 63;
y = 9;

From here, we can determine that z = 27 & x = 4;

Since we are asked to find the product, we can find xyz: 4 * 9 * 27 = 972;

Ans is (B).

a+b+c=40 (1)
Let a<b<c
Therefore,c=3b and a=c-23.
Replacing, a and b in (1)
we get c=27
a=4 and b=9
a*b*c=4*9*27=972
Ans. B

When you guys post questions from the Official Review...How do you know the difficulty level?

The difficulty is crazy low, but its a 600 Level Question?


and yah, I got 972 as well

A+B+C=40

A+B+3B=40
3B-23+B+3B=40
7B=63

\(\frac{7B}{7}\) \(=\) \(\frac{63}{7}\)

\(B=9\)

The rest is history

I represent my three numbers as three different variables because it shows how they can be non or interrelated when I started the problem, I had no idea about the relationship(s) of the numbers and never assume they are "n" related, besides being integers.

The sum of three integers is 40. The largest integer is 3 times the middle integer, and the smallest integer is 23 less than the largest integer. What is the product of the three integers?

(A) 1,104
(B) 972
(C) 672
(D) 294
(E) 192

Let the numbers be \(x_1\), \(x_2\) and \(x_3\)

Given that, \(x_3 = 3*x_2\) Or, \(x_2 = x_3/3\) & \(x_1 = x_3 -23\)

Since,\(x_1+ x_2 + x_3 = 40\)

Or,\(x_3-23 +\frac{x_3}{3} + x_3=40\)

Or, \(3* x_3+x_3 + x_3*3=(40 + 23)*3\)

Or, \(7x_3=189\)

Or, \(x_3=27\)

Substituting the value of \(x_3\), \(x_1= x_3-23=4\) & \(x_2=x_3/3=9\)

So, \(x_1*x_2*x_3=4*9*27=972\)

Answer: (B)

L + M + S = 40
3M + M + 3M - 23= 40
7M=63, M=9 -> L=27 -> S=4 ---> 4*27*9 ~ 100*9 = 900 (B)

Hi All,

This prompt is built on a few Number Properties, so you can actually solve it with a bit of 'brute force' math (without the need of any lengthy equations or calculations:

We're given a few facts to work with:
1) The SUM of 3 integers is 40
2) The largest integer = 3(middle integer)
3) The largest integers = 23 + smallest integer

We're asked for the PRODUCT of the three integers.

First, notice how all 3 answers are POSITIVE, that means that our 3 integers are likely all positive. From the given information, we can see that the largest integer will be considerably bigger than each of the other two.

From the first fact, we know that the largest integer is 3 TIMES the middle integer, so the largest integer MUST be a multiple of 3. Here, we can start brute-forcing the work by 'playing around' with multiples of 3...

IF...
the largest number is 21, then the other two numbers are 7 and -2; the sum is 26, which is TOO LOW. Let's try something bigger....

IF....
the largest number is 24, then the other two numbers are 8 and 1; the sum is 33, which is TOO LOW. We have to go bigger...

IF....
the largest number is 27, then the other two numbers are 9 and 4; the sum IS 40, so these MUST be the 3 numbers

The answer to the question will be the product of 27, 9 and 4. You can organize this math in variety of ways, but you might find it easiest to look at it this way....

(27x4)(9) = (108)(9) --> something MORE than 900 but LESS than 1,000. You might also notice that since one of the numbers is 9, the product MUST be a multiple of 9 (and there's only one answer that fit's THAT description).

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Here is my approach =>
let x,y,z be the three integers in decreasing order
here x+y+z=40
x+x/3+x-23=40
=> 7x=63*9=> x=27
so y=27/3=9
z=27-23=4
hence the product = 27*9*4
also as the only option that is a multiple of three is B => Smash it

We can create the following 3 equations in which a = the smallest integer, b = the middle integer, and c = the largest integer:

a + b + c = 40

and

c = 3b

c/3 = b

and

c - 23 = a

Thus:

c - 23 + c/3 + c = 40

2c + c/3 = 63

Multiplying by 3, we have:

6c + c = 189

7c = 189

c = 27, so a = 27 - 23 = 4 and b = 27/3 = 9.

Thus, the product of the 3 integers is 4 x 9 x 27 = 972.

Answer: B

Hi,

How do we know that the product ((3x-23)*x*3x) is a multiple of 9?

Doesn't the -23 mean that the product is whatever -23?

Hi Gradus,

Once you know that one of the three integers is 9 (in this case, that's the "middle number") which is a multiple of 9.... then the product of that number and two other INTEGERS will ALSO be a multiple of 9. While you're correct that the "-23" means that (3X - 23) is NOT a multiple of 9, that does not matter - again, since we're multiplying two integers by 9, the resulting product WILL be a multiple of 9.

GMAT assassins aren't born, they're made,
Rich

x,y,z be the three numbers in increasing order.

z=3y
x=z-23
x=3y-23

x+y+z=40
3y-23+y+3y=40
y=9
z=27
x=4

Product of x,y, and z = 972

Option B.

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