GMAT Club Olympics 2024 (Day 5): What is the largest 4-digit number th

Question

What is the largest 4-digit number that can be added to 7,700 for the sum to be divisible by 22, 25, and 32?

A. 1,100
B. 2,200
C. 7,700
D. 8,800
E. 9,900

Bunuel · Accepted Answer

GMAT Club Official Explanation:

For a number to be divisible by 22, 25, and 32, it should be divisible by the least common multiple of these numbers. The LCM of 22 (2 * 11), 25 (5^2), and 32 (2^5) is 2^5 * 5^2 * 11 = 8,800.

Thus, we need the largest 4-digit number n such that 7,700 + n is a multiple of 8,800. The smallest positive number we could add to 7,700 to make the result divisible by 8,800 is 8,800 - 7,700 = 1,100. On the other hand, the largest 4-digit number that can be added to 7,700 to achieve this is 1,100 + 8,800 = 9,900. So, we get 7,700 + n = 7,700 + (1,100 + 8,800) = 7,700 + 9,900.

Answer: E

Purnank · Answer

I think E.
Given -  What is the largest 4-digit number that can be added to 7,700  for the sum to be divisible by 22, 25, and 32?
Lets first calculate LCM of 22, 25, 32 
22=11*2
25= 5^2 
32= 2^5 
Therefore,
LCM of 22, 25, 32 = ( 5^2  *  2^5  * 11) = 8800
Now,
7700 + x = 8800k
for k=1, x=8800 - 7700 = 1100 
But we need larger value so lets check again for higher k value.
k=2, x=8800*2 - 7700 = 17600 - 7700 = 9900
Matches with given option plus it is highest value that option contains so
 Answer is E.

Kinshook · Answer

What is the largest 4-digit number that can be added to 7,700 for the sum to be divisible by 22, 25, and 32?

22 = 2*11
25 = 5*5
32 = 2^5

For the sum to be divisible by 22, 25 & 32, it should be divisible by LCM(22,25,32) = 2^5*5^2*11 = 8800

The largest 4-digit number that can be added to 7,700 for the sum to be divisible by 22, 25, and 32 = 2*8800 - 7700 = 9900

IMO E

Oppenheimer1945 · Answer

all numbers N+7700 are multiple of 11x2, 25 , We only need to check if N+7700 has a factor of 2^5

Only A & E are. Hence largest no=9900

E)

Evronredor · Answer

What is the largest 4-digit number that can be added to 7,700 for the sum to be divisible by 22, 25, and 32?

For the number to be divisible by 22,25 and 32, it should have the following factors -  2^5, 5^2 and 11

Backsolving using the options, starting with the middle option i.e. C

C) 7700 - 7700+7700 = 15400

factors of 15400 => 2^3, 5^2, 7 and 11

D) 8800 - 8800+7700= 16500

factors of 16500 => 2^2, 3, 5^3 and 11

E) 9900 - 9900+7700= 17600

factors of 17600 =>  2^5, 5^2 and 11

Answer: E

captain0612 · Answer

Choice E

Largest 4 digit integer that can be added to 7700 so that it is divisible by 22, 25, 32 is?

Step 1: For the new sum to be divisible by 22, 25 and 32, the LCM of the 3 numbers should be divide the sum
Hence LCM of 22, 25 , 32 is
22 : 2*11
25 : 2*5
32 : 2^5

LCM : 2^5 * 5 * 11 = 8800

Since, 8800 > 7700 a multiple of 8800 will divide the sum

8800 * 2 = 17600

17600 - 7700 = 9900

9900 is the largest number that can be added to 7700 so that the sum is divisible by all 3 numbers : 22, 25, 32

A_Nishith · Answer

To find the largest 4-digit number that can be added to 7,700 so that the sum is divisible by 22, 25, and 32, we need to determine the least common multiple (LCM) of these numbers and then find the largest 4-digit number satisfying this condition.

Step 1:  Calculate the LCM of 22, 25, and 32
First, find the prime factorizations:
22=2×11
25=5^2
32=2^5

The LCM is the highest power of each prime that appears in any of the factorizations:

Highest power of 2: 2^5
Highest power of 5: 5^2
Highest power of 11: 11
Therefore, the LCM is:
LCM = (2^5)×(5^2)×11

Calculate this value:
2^5 =32
5^2 = 25
32×25= 800
800×11= 8800

So, the LCM of 22, 25, and 32 is 8800.

Step 2:  Find the largest 4-digit number that, when added to 7,700, makes the sum divisible by 8800
Let x be the largest 4-digit number such that: 
=> 7700+x≡0(mod8800)

We need to find x such that 7700+x is a multiple of 8800: 7700+x=8800k
for some integer k.

First, solve for x:
x=8800k−7700

We need x to be a 4-digit number.  Thus: 1000≤x≤9999

We now  find the largest x within this range: 8800k−7700 ≤ 9999

Rearrange and solve for k: 
8800k≤17699

k  ≤ 17699/8800 ≈  2.01

Since k must be an integer, the largest possible value of k is 2.

Substitute k=2 into the equation for x:
x=8800×2−7700
x=17600−7700
 x=9900

Thus, the largest 4-digit number that can be added to 7700 so that the sum is divisible by 22, 25, and 32 is: 9900 
 
Answer: E

aviraj1703 · Answer

L.C.M of (22, 25, 32) = 2^5 x 5^2 x 11
So, resulting number must contain 2^5 x 5^2 x 11 in its factors to be divisible by 22, 25 and 32.

Now, 7700 = 2^2 x 5^2 x 7 x 11 and let say we are adding x which is 4 digits number to the given number.
As resulting number must be multiple of 8800, added number has to be multiple of 1100.
According to options x is clearly having 2^2 x 5^2 x 11 as factors and let assume remaining factors as y.

Therefore, (2^2 x 5^2 x 7 x 11) + (2^2 x 5^2 x 11 x y) = 2^2 x 5^2 x 11 x (7 + y)
For 2^2 x 5^2 x 11 x (7 + y) to be divisible by 2^5 x 5^2 x 11, (7 + y) must be having 8 as a factor.
7 + y >= 8 => y >= 1

Here, y can not be even else it 7 + y will be odd number and we won't get answer. Also y can not be greater than 9 else added number will be having 5 digits.

According to options(1, 2, 7, 8, 9) y can be 1 or 9 and as we won't the largest number we will go with 9.

Answer: E. 9,900

DG1989 · Answer

What is the largest 4-digit number that can be added to 7,700 for the sum to be divisible by 22, 25, and 32?

A. 1,100
B. 2,200
C. 7,700
D. 8,800
E. 9,900

Solution:  To find the largest 4-digit number that can be added to 7,700 such that the sum is divisible by 22, 25, and 32, we need to first find the least common multiple (LCM) of these numbers.

LCM of 22, 25, and 32 = LCM (2 * 11,  5^2 ,  2^5 )
=  2^5  *  5^2  * 11
= 8800

Let x be the 4-digit number to be added to 7700.
Then, 7700 + x should be divisible by 8800
This means, 7700 + x = 8800k               (for some integer k)
x = 8800k - 7700

For k = 1
x = 8800 - 7700
x = 1100

For k = 2
x = 17600 - 7700
x = 9900

For k = 3 or more, x won't be a 4-digit number
Thus, the largest 4-digit number that can be added to 7700 = 9900

The correct answer is E

Spectre26 · Answer

7700 = 7*11*25*4

Let's check the options:

A.  1100 = 11*25*4 
 7700 + 1100 = 8*11*25*4 , This is divisible by all three  (22, 25, 32)

B.  2200 = 2*11*25*4 
 7700 + 2200 = 9*11*25*4 , NOT divisible by  32

C.  7700 = 7*11*25*4 
 7700 + 7700 = 2*7*11*25*4 , NOT divisible by  32

D.  8800 = 8*11*25*4 
 7700 + 8800 = 15*11*25*4 , NOT divisible by  32

E.  9900 = 9*11*25*4 
 7700 + 9900 = 16*11*25*4 , This is divisible by all three  (22, 25, 32) 
Also this is the largest amongst the five options

Correct Option: E

GuadalupeAntonia · Answer

N= largest 4-digit number divisible by 22, 25, and 32
GCM (22,25,32)=2200

so N must be a multiple of 2200 such that 7700+x=2200*(m) 
 m must be an integer such that
2200*(m)=< 7700+9999
2200*(m)=< 17,699
m=< 8.somenthing

let be m=8

7700+x=2200*(8) 
7700+x=17,600
x=17,600-7700= 9,900

AND letter  E

HarishD · Answer

What is the largest 4-digit number that can be added to 7,700 for the sum to be divisible by 22, 25, and 32?

LCM of 22, 25 and 32 is 8800

number: 8800*n is divisible by 22, 25 and 32
 
n=1 number: 8800 to add: 1100
n=2 number: 17600, to add: 9900
n=3 number: 26400, to add: 18700 (more than 4 digits)

Therefore answer is 9900 (E)

PK1 · Answer

22 = 2*11
25= 5* 5
32= 2*2*2*2*2
LCM of (22,25,32)=32*11*5*5=160*5*11=800*11=8800
so any number divisible by 22, 25, 32 needs to be of the form 8800k

7700+x= 8800
so x =1100

7700+x=17600 (8800*2)
x=17600-7700= 9900

largest 4 digit is hence 9900

E is the answer

Sof22 · Answer

The number is divisible by 22 if it is divisible by 2 and 11
The number is divisible by 25 when its last two digits form such combinations: 00, 25, 50 or 75
The number is divisble by 32 when it is divisible by 2 and 16

Let's start directly with 9,900+7,700 = 17600

17600 is divisable by 2 and 11 (1+6+0=7+0), so it is divisable by 22
17600 is divisable by 25 
17600 is divasable by 2 and 16, so it is divasable by 32

So, E is the correct answer

Gemmie · Answer

Question: Find the largest N so that N + 7700 is divisible by 22, 25, and 32

(i) 7700 = 11 * 7 * 4 * 25

=> 7700 is divisible by 22 and 25

=> N is divisible by 22 and 25

In fact, all answer choices are divisible by 22 and 25

(ii) 7700 = 4 * 77 * 25

77 = 8x + 5
25 = 8y + 1

=> 7700 = 4 * (8x + 5) * (8y + 1) = 4 * (8m + 5) = 32m + 20

=> N must leave a remainder of 12 when divided by 32

9900 = 32 * 309 + 12

==> max N = 9900

OmerKor · Answer

Ok, We need to fight back if we want to WIN.
Green team, it's time to wake up!
Fifth Day, Here we go:
Let's get started with our explanation for this topic:

Glance: the Question&Answer Choices:
Question: We have here kind of Factors question.
Answer Choices: seperate from each other. multiples of 11. it implys that we be able to work backward here.

Reading and Understanding the question:
Given: [7700 + X (one of our answer choices)][/22*25*32]
[?]: we are trying to find the biggest number of X (one of our answer choices) which will divide by these 3 numbers.
22 = 2 * 11
25 = 5^2
32 = 2^5
All together are: 2^6 * 5^2 * 11

Plan
We can work backward here. since we need the biggest number. we can start with 9,900.

Solve:
7,700 + 9,900 = 17,600
17,600 = 2^6 * 5^2 * 11
we can see that this is exactly matching our prethinking of the 3 numbers (22*25*32): 2^6 * 5^2 * 11.
So this is the biggest number we can add that will divide with these numbers.
Answer Choice E is the answer

THE END
I hope you liked the explanation, I have tried my best here.
Let me know if you have any questions about this question or my explanation.

THE END I hope you liked the explanation, I have tried my best here. Let me know if you have any questions about this question or my explanation. https://cdn.gmatclub.com/cdn/files/forum/images/smilies/1f44f-1f3fb.png

0ExtraTerrestrial · Answer

For any number to be divisible by 22, 25 and 32, the least condition it should satisfy is the it must be divisible by the LCM of 22, 25 and 32.
22 = 2 * 11
25 = 5^2
32 = 2^5

Thus LCM(22, 25, 32)= 2^5 * 5^2 * 11 = 8800.

If we were asked what should be added to 7700 for the sum to be a four digit number and divisible by 22, 25 and 32 then 8800 being largest such possible number, the answer would'be been 1100. But here we are asked, what is the largest 4 digit number that can be added to 7700 so that the sum is divisible by 22, 25 and 32. Here we can write such sum as 8800k with k being an integer and difference as 8800k - 7700.

Thus 8800k-7700 is a 4 digit number and largest possible such integer.
8800k - 7700 = 1100(8k-7)
Since 1100 is already 4 digits, 8k-7 must be a single digit, largest single digit is 9, since 9 can be written as 8k-7, the answer is 9*1100= 9900.

Posted from my mobile device

Lizaza · Answer

Starting with the desired result, we need to find the smallest number that's divisible by all three values:
 22 = 11*2 
 25 = 5^2 
 32 = 2^5 
Hence, the number we seek must contain  2^5*5^2*11 = 8800

Obviously, if we add 1100 to 7700, we get 8800 immediately. But can we add sth bigger?

8800*2 = 17600 
  17600 - 7700 = 9900

We still  have this number among the options  - therefore,   the right answer is E.

d_patel · Answer

We need to find largest 4-digit number that can be added to 7,700 for the sum to be divisible by 22, 25, and 32.

Remember that if a number is divisible by 22, 25, and 32 then it must be divisble by LCM of 22, 25, and 32.

22 = 2 * 11 
 25 = 5^2 
 32 = 2^5

Higher power of 2 is 5, 5 is 2 and 11 is 1

LCM(22, 25, 32) =  2^5 * 5^2 * 11^1

Now as we need to find largest number, let's start with largest option available and then decrease it if option doesn't work out.

Here largest number =  9900 
New number =  9900 + 7700   =  17600

Let's find prime factors of  17600 ,
 17600 = 176 * 100 
=  16 * 11 * 25 * 4 
=  2^7 * 5^2 * 11^1

And LCM was  2^5 * 5^2 * 11^1 .

So, 17600 is divisible by LCM.
So, 17600 is divisible by 22, 25 and 32.

Select  option-E  as answer.

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