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\(&\) represents the tens digit in integer \(A = 1543&2\) . What is \(&\) ? 1. \(A\) is divisible by 9 2. \(A\) is divisible by 4 Source: GMAT Club Tests  hardest GMAT questions S1 is sufficient. \(1+5+4+3+&+2\) must be divisible by 9. This is possible only when \(& = 3\) . S2 is not sufficient. \(&2\) must be divisible by 4. This is possible when \(& = 1\) or \(& = 3\) . I realize this is a pretty rudimentary question but would somebody please explain why all the digits are added together to determine divisibility by 9 for statement 1, but for statement 2, only the first two digits are added to determine divisibility. The answer is just not obviously clear to me.



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Re: M22 #10 [#permalink]
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03 Oct 2008, 10:20
Hey answer is pretty simple. !! there are few rules for division you need to understand before answering such kind of questions. Rule I. divisible by "2"  any no. will be considered divisible by two if the last digit of that no. is divisible by 2.
ex XXX4  this no. will always be divisible by 2
Rule II. for numbers "3" and "9"
if sum of the digits of number is divisible by "3" or "9" then the number is considered divisible by "3" or "9" consider no. 123456 , sum of digits = 1+2+3+4+5+6=21 , now 21 is divisible by 3 only not by 9. therefore 123456 is divisible by 3 only.
Rule III divisible by "4" for any no. to be divisible by 4 , its last two digits should be divisible by 4. eg. X32 will always be divisible by 4 for any value of X. You can try this exercise yourself



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Re: M22 #10 [#permalink]
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09 Oct 2008, 02:39
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Sum of digits should be divisible by 9 in order for a number to be divisible by 9.... so add the digits.



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Re: M22 #10 [#permalink]
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04 Feb 2010, 22:24
1.A is divisible by 9 2. A is divisible by 4
since & is in s 10th digit it can accommodate only one digit which is divisible by 9 so choice 1 alone sufficient
2) & can have 2 possible digits which is divisible by 4 so choice 2 alone is insufficient.
so ans is A



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Re: M22 #10 [#permalink]
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07 Feb 2010, 23:08
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A small tip: Divisibility rule for 2^n  Last n digits should be divisible by the number
For example for 2  check last 1 digit for 4  check last 2 digits for 8  check last 3 digits etc.



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Re: M22 #10 [#permalink]
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08 Feb 2010, 06:02
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@meetnaren Thanks for the valuable tip, but it would be great if u could explain it a little more, so that it is easy to understand. At first, I didn't get but then I used some random numbers & divided them to verify and it did work ( maybe I've got dull a bit )
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Re: M22 #10 [#permalink]
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19 Mar 2010, 08:40
Rules are like following: 2: last digit should be divisible by 2. 3: sum of the digits should be divisible by 3. 4: last 2 digits should be divisible by 4. 5: last digit should be 0 or 5. 6: last digit should be divisible by 2 and sum of the digits should be divisible by 3. 7: given below 8: last 3 digits should be divisible by 8. 9: sum of the digits should be divisible by 9. Divisibility by 7: Take the last digit in a number. Double and subtract the last digit in your number from the rest of the digits. Repeat the process for larger numbers. Example: 287 (Double the 7 to get 14. Subtract 14 from 28 to get 7 which is divisible by 7 and we can now say that 287 is divisible by 7. Among these 7 is a bit tricky.
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Re: M22 #10 [#permalink]
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10 Feb 2011, 06:40
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A = 1543&2. What is &?
1  A is divisible by 9. That means 1+5+4+3+2+& must be divisible by 9. Now 1+5+4+3+2 is 15. Although, in general & can any integer value from 0  9, but it can only take a value that makes A a multile of 9. When &=3, A = 18 > A is divisible by 9 we need 9 more to reach next multiple of 9, but since we have already used 3 to reach 18 (15 + 3 = 18), we need 12 to reach next multiple of 9, i.e 27. (15 + 12 = 27) Because & = 12 is not possible, we can only use & = 3. Statement 1 is sufficient.
2  For a number to be divisible by 4, its last two digits must be divisible by 4 (i.e a multiple of 4). Considering this rule, 154312 should be divisible by 4. 154352 should also be divisible by 4. Because we have found two values for &, this statement is not sufficient.
Thus answer is A.
Hope my explanations helps...



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Re: M22 #10 [#permalink]
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13 Feb 2012, 10:58
@meetnaren or @AtifS  can you explain the rule? seems like a rule that could be really helpful. Thks



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Re: M22 #10 [#permalink]
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13 Feb 2012, 11:16
T740qc wrote: @meetnaren or @AtifS  can you explain the rule? seems like a rule that could be really helpful. Thks Divisibility Rules 2  If the last digit is even, the number is divisible by 2. 3  If the sum of the digits is divisible by 3, the number is also. 4  If the last two digits form a number divisible by 4, the number is also. 5  If the last digit is a 5 or a 0, the number is divisible by 5. 6  If the number is divisible by both 3 and 2, it is also divisible by 6. 7  Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7. 8  If the last three digits of a number are divisible by 8, then so is the whole number. 9  If the sum of the digits is divisible by 9, so is the number. 10  If the number ends in 0, it is divisible by 10. 11  If you sum every second digit and then subtract all other digits and the answer is: 0, or is divisible by 11, then the number is divisible by 11. Example: to see whether 9,488,699 is divisible by 11, sum every second digit: 4+8+9=21, then subtract the sum of other digits: 21(9+8+6+9)=11, 11 is divisible by 11, hence 9,488,699 is divisible by 11. 12  If the number is divisible by both 3 and 4, it is also divisible by 12. 25  Numbers ending with 00, 25, 50, or 75 represent numbers divisible by 25. For more on Number Theory check this: mathnumbertheory88376.htmlHope it helps.
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Re: M22 #10 [#permalink]
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13 Feb 2012, 13:03
Bunuel wrote: T740qc wrote: @meetnaren or @AtifS  can you explain the rule? seems like a rule that could be really helpful. Thks Divisibility Rules 2  If the last digit is even, the number is divisible by 2. 3  If the sum of the digits is divisible by 3, the number is also. 4  If the last two digits form a number divisible by 4, the number is also. 5  If the last digit is a 5 or a 0, the number is divisible by 5. 6  If the number is divisible by both 3 and 2, it is also divisible by 6. 7  Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7. 8  If the last three digits of a number are divisible by 8, then so is the whole number. 9  If the sum of the digits is divisible by 9, so is the number. 10  If the number ends in 0, it is divisible by 10. 11  If you sum every second digit and then subtract all other digits and the answer is: 0, or is divisible by 11, then the number is divisible by 11. Example: to see whether 9,488,699 is divisible by 11, sum every second digit: 4+8+9=21, then subtract the sum of other digits: 21(9+8+6+9)=11, 11 is divisible by 11, hence 9,488,699 is divisible by 11. 12  If the number is divisible by both 3 and 4, it is also divisible by 12. 25  Numbers ending with 00, 25, 50, or 75 represent numbers divisible by 25. For more on Number Theory check this: mathnumbertheory88376.htmlHope it helps. Thank you for this tip. I had learned all the divisibility rules up to 9 but had not seen the rule for 11. Very helpful.



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Re: M22 #10 [#permalink]
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13 Feb 2013, 09:40
snowy2009 wrote: \(&\) represents the tens digit in integer \(A = 1543&2\) . What is \(&\) ? 1. \(A\) is divisible by 9 2. \(A\) is divisible by 4 Source: GMAT Club Tests  hardest GMAT questions S1 is sufficient. \(1+5+4+3+&+2\) must be divisible by 9. This is possible only when \(& = 3\) . S2 is not sufficient. \(&2\) must be divisible by 4. This is possible when \(& = 1\) or \(& = 3\) . I realize this is a pretty rudimentary question but would somebody please explain why all the digits are added together to determine divisibility by 9 for statement 1, but for statement 2, only the first two digits are added to determine divisibility. The answer is just not obviously clear to me. \(@\) represents the tens digit in integer \(A = 1543@2\). What is the value of \(@\) ? (1) \(A\) is divisible by 9. In order an integer to be divisible by 9, the sum of its digit must be divisible by 9. Hence \(1+5+4+3+@+2=15+@\) must be divisible by 9, and since @ is a digit then it can only equal to 3. Sufficient. (2) \(A\) is divisible by 4. In order an integer to be divisible by 4, its last two digits must be divisible by 4. Hence \(@2\) must be divisible by 4, which is possible if \(@=1\), \(@=3\), \(@=5\), \(@=7\), or \(@=9\). Not sufficient. Answer: A.
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Re: M22 #10 [#permalink]
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13 Feb 2013, 11:03
We will have to check the divisibility rules for both statements: For S1 to be true &=3. hence S1 is sufficient For S2 to be true & =1/3 (as last two digits have to be divisible by 4). hence S2 is not sufficient. Thus answer is A



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Re: M22 #10 [#permalink]
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27 Mar 2014, 12:28
Bunuel wrote: T740qc wrote: @meetnaren or @AtifS  can you explain the rule? seems like a rule that could be really helpful. Thks Divisibility Rules 2  If the last digit is even, the number is divisible by 2. 3  If the sum of the digits is divisible by 3, the number is also. 4  If the last two digits form a number divisible by 4, the number is also. 5  If the last digit is a 5 or a 0, the number is divisible by 5. 6  If the number is divisible by both 3 and 2, it is also divisible by 6. 7  Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7. 8  If the last three digits of a number are divisible by 8, then so is the whole number. 9  If the sum of the digits is divisible by 9, so is the number. 10  If the number ends in 0, it is divisible by 10. 11  If you sum every second digit and then subtract all other digits and the answer is: 0, or is divisible by 11, then the number is divisible by 11. Example: to see whether 9,488,699 is divisible by 11, sum every second digit: 4+8+9=21, then subtract the sum of other digits: 21(9+8+6+9)=11, 11 is divisible by 11, hence 9,488,699 is divisible by 11. 12  If the number is divisible by both 3 and 4, it is also divisible by 12. 25  Numbers ending with 00, 25, 50, or 75 represent numbers divisible by 25. For more on Number Theory check this: mathnumbertheory88376.htmlHope it helps. Dear Bunuel As 0 is the first multiple of every number than will it not be divisible by every number. thanks Sid



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Re: M22 #10 [#permalink]
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28 Mar 2014, 01:50
sidpopy wrote: Bunuel wrote: T740qc wrote: @meetnaren or @AtifS  can you explain the rule? seems like a rule that could be really helpful. Thks Divisibility Rules 2  If the last digit is even, the number is divisible by 2. 3  If the sum of the digits is divisible by 3, the number is also. 4  If the last two digits form a number divisible by 4, the number is also. 5  If the last digit is a 5 or a 0, the number is divisible by 5. 6  If the number is divisible by both 3 and 2, it is also divisible by 6. 7  Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7. 8  If the last three digits of a number are divisible by 8, then so is the whole number. 9  If the sum of the digits is divisible by 9, so is the number. 10  If the number ends in 0, it is divisible by 10. 11  If you sum every second digit and then subtract all other digits and the answer is: 0, or is divisible by 11, then the number is divisible by 11. Example: to see whether 9,488,699 is divisible by 11, sum every second digit: 4+8+9=21, then subtract the sum of other digits: 21(9+8+6+9)=11, 11 is divisible by 11, hence 9,488,699 is divisible by 11. 12  If the number is divisible by both 3 and 4, it is also divisible by 12. 25  Numbers ending with 00, 25, 50, or 75 represent numbers divisible by 25. For more on Number Theory check this: mathnumbertheory88376.htmlHope it helps. Dear Bunuel As 0 is the first multiple of every number than will it not be divisible by every number. thanks Sid Yes, zero is divisible by EVERY integer except zero itself, since 0/integer=integer (or, which is the same, zero is a multiple of every integer except zero itself).
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Re: M22 #10 [#permalink]
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21 Apr 2014, 05:29
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1) A is divisible by 9 > 1+5+4+3+2+& = 15+& is divisible by 9 (0=<&=<9) > &=3 > sufficient 2) A is divisible by 4 > &2 is divisible by 4. &=3 and &=1 are both ok > insufficient
Choose A











