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hey bb and bunuel!
can I apply to colleges with different email ID than the one used in official mba.com, from which I have sent scores to those colleges

I have college application wavier from different email id, will it create any issue?

Your mba.com email has ZERO impact on anything.

We are given the expression \( 10^{25} - 560 \) and asked to determine which of the following numbers (11, 8, 5, 4, or 3) **does not divide** the result.

Let’s evaluate the divisibility of \( 10^{25} - 560 \) by each of these numbers.

### **Step 1: Divisibility by 5**
Any number \( 10^n \) (where \( n \) is a positive integer) is divisible by 5 since it ends in zero. So, \( 10^{25} \) is divisible by 5, and 560 is also divisible by 5 because \( 560 \div 5 = 112 \).

Thus, \( 10^{25} - 560 \) is divisible by 5.

- **Not the answer.**

### **Step 2: Divisibility by 4**
We check divisibility by 4 using the last two digits of a number. The last two digits of \( 10^{25} \) are "00", and the last two digits of 560 are "60".

Thus, \( 10^{25} - 560 \) ends in 40, which is divisible by 4.

- **Not the answer.**

### **Step 3: Divisibility by 8**
We check divisibility by 8 using the last three digits of the number. The last three digits of \( 10^{25} \) are "000", and the last three digits of 560 are "560".

Now, \( 1000 - 560 = 440 \), and 440 is divisible by 8 because \( 440 \div 8 = 55 \).

Thus, \( 10^{25} - 560 \) is divisible by 8.

- **Not the answer.**

### **Step 4: Divisibility by 11**
To check divisibility by 11, we apply the alternating sum rule. For a number to be divisible by 11, the alternating sum of its digits must be divisible by 11.

The alternating sum of the digits of \( 10^{25} \) is \( 1 - 0 + 0 - 0 + \dots + 0 = 1 \), and the alternating sum of the digits of 560 is \( 5 - 6 + 0 = -1 \).

Thus, the alternating sum of \( 10^{25} - 560 \) is \( 1 - (-1) = 2 \), which is **not divisible by 11**.

Therefore, \( 10^{25} - 560 \) is **not divisible by 11**.

- **Answer: 11**

### **Step 5: Divisibility by 3**
To check divisibility by 3, we sum the digits of the number. If the sum of the digits is divisible by 3, the number is divisible by 3.

The sum of the digits of \( 10^{25} \) is 1, and the sum of the digits of 560 is \( 5 + 6 + 0 = 11 \). So, the sum of the digits of \( 10^{25} - 560 \) is \( 1 - 11 = -10 \), and the absolute value is 10, which is **not divisible by 3**.

However, there can be a calculation mix-up, so let’s verify step-by-step manually

L

thankyou bb!! I was in doubt and heard mixed opinion, clear now!

Irrespective of the approach you use, your answer in the end has to be correct....you are mentioning 11 as the answer which is wrong.....You might see Bunuel has already shared the link for the answer.......how I approached the problem is ...10^25 - 564 --> I used 10^5 as an eg - 560 = 99440.....did a Prime factorization of 440 --> 2 * 2 *2 * 5 * 11 so that tells you that the part 440 comprise 2, 4, 5, 6, 11 as factors......so for the no ending with 440 U can conclude it’s divisible by 4, 5 and 8......now for the 1st part 10^6 -560 --> 999440 10^7 - 560 --> 9999560........for odd no of 9’s eg 999440 no won’t be divisible by 11 but for 10^25 - 560 --> thr must be 22 9’s440 --> even no of 9’s means no is divisible by 11......only option remaining is 3 which shd be the answer.......logic for non-divisibility by 3 even no of 9’s (22 *9 + 4 + 4) = sum of digits is an even no so not divisible by 3

Hello. How do i solve this
Which of the following must be an integer if p is a positive integer and (3/p + 5/p) is also an integer?

Options?

8=pk so p can be 1,2,4,8

A. p/4
B. 24/p
C.12/p
D. 30/p
E. None

Ok ok got it. Thanks

24/p=3k which is always an integer

Hi ^^ Just a short question did you have to wait some hours after you bought a sub to get the tests unlocked? Somehow they are still greyed out. Even after i used the links.

Hello . How do I solve this?

A man has twice as many 25-paisa coins as he has 10-paisa coins now. If he had taka 1.50 more in 2 hand and replaced it with all 10-paisa coins, he would have twice as many 10-paisa coins as he has 25-paisa coins. How many 25-paisa coins does he have now?
a.5

b. 10

c. 15

d. 20

e. 25

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Hi Fariha.....2nd sentence is a bit messed up the way U got the Q and also we shd know taka to paisa conversion....if taka is Rupees then 1 Rs = 100 paisa...if taka is Bangladesh’s currency then I think it’s 1 taka = 120 paisa.....however the logic shd be comething like assume 10 paisa coins to be x then 25 paisa coins wd be 2x (twice the no) ......1.5 taka replaced with 10 paisa coins assuming 1 taka = 100 paisa wd require (15) 10 paisa coins ....so total no of 10 paisa x + 15 = 2 * (2x) [total no of 25 paisa] --> x + 15 = 4x --> x =5 so no of 5 paisa coins =5 no of 25 paisa coins wd be 2*5 = 10 so option B

hello I redeemed points 2 days back still didn’t got the gift. Can anyone tell how much time it take.?

Hello. How do i solve this?
The remainder when m + n is divided by 12 is 8, and the remainder when m – n is divided by 12 is 6. If m > n, then what is the remainder when mn divided by 6?
A)1
B)2
C)3
D)4
E)5

Discussed in detail here: https://gmatclub.com/forum/the-remainde ... 62707.html

Hope it helps.

Hi! I am trying out the forum quiz but I would like to avoid the official prep test questions because I am yet to take them. I’ve seen in another thread that I should be able to filter them out but I dont see a "gmat prep" in the filter options

How should I filter out these questions?