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How many threedigit integers between 310 and 400, exclusive [#permalink]
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11 Aug 2008, 13:26
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How many threedigit integers between 310 and 400, exclusive, are divisible by 3 when the tens digit and the hundered digit are switched? A. 3 B. 19 C. 22 D. 30 E. 90
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Re: Factor problem [#permalink]
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11 Aug 2008, 13:40
I think its 5) 90
Numbers b/w 310  400, ie 312 to 399 (coz they would be divisible by 3) = 87
"are divisible by 3 when the tens digit and the hundered digit are switched"
330, 331, 332, 333, 334, 335, 336, 337, 338, 339  can only use these numbers to satisy the question. Now, these divisible by 3 would be 10/3 = 3.33 i.e 3
So, 87+3 = 90?
whats OA?



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Re: Factor problem [#permalink]
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Updated on: 11 Aug 2008, 13:48
rnemani wrote: How many threedigit integers between 310 and 400, exclusive, are divisible by 3 when the tens digit and the hundered digit are switched? 1) 3 2) 19 3) 22 4) 30 5) 90 Totally, lost on the solutions and the approach, please help Alright, first off you have to remember the divisibility rule for 3: If all the numbers added together are divisible by 3 then the number itself is divisible by 3. Ex: 312, 3+1+2=6 (6 is divisible by 3, hence 312 is divisible by 3). Now, start listing all the number after 310 that are divisible by 3 (because of the divisibility rule, switch the tens and hundreds place should make no difference). 312 315 318 321 324 327 330 333 336 339 342 345 348 ..1 ..4 ..7 ..0 ..3 ..6 ..9 pattern repeats till 399. you can count there are 10 numbers divisible by 3 between 310340 and since the pattern repeats there should be 10 between 340370 and 370400. hence, there are 30 numbers.
Originally posted by djsbee on 11 Aug 2008, 13:44.
Last edited by djsbee on 11 Aug 2008, 13:48, edited 2 times in total.



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Re: Factor problem [#permalink]
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11 Aug 2008, 13:44
rnemani wrote: How many threedigit integers between 310 and 400, exclusive, are divisible by 3 when the tens digit and the hundered digit are switched? 1) 3 2) 19 3) 22 4) 30 5) 90 Totally, lost on the solutions and the approach, please help Answer is 30 No of digits between 310 and 400 (exclusive) = 4003101 = 89 No Intergers divisble by 3 =~ 89/3 ~ 30 Divsiblivity rule for 3 is Sum of all digits must be divisble by 3. It doesn't matter if hundred digit and tens digits are switched. e.g 372 is divisble by 3 (becae 3+7+2= 12 (divisble by 3)) switch digits 732 also divisble by 3 (becae 7+3+2= 12 (divisble by 3))
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Re: Factor problem [#permalink]
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12 Aug 2008, 00:08
IMO 30
HERE IT GOES..
IF WE SEE ALL THE THREEDIGIT NOS DIVISIBLE BY 3..THEN ALL OF THEM SATISFY THE ABOVE CONDITION..(SWAPPING TENS DIGIT WITH HUNDREDS DIGIT) BECAUSE THE SUM OF THE DIGITS REMAINS THE SAME AND THAT IS ALL WE REQUIRE TO SEE IF A NO. IS DIV BY 3 OR NOT
SO,WE NEED TO FIND ONLY THE NO. OF FACTORS OF 3 BETWEEN 310 & 400
133  102 + 1 =30



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Re: Factor problem [#permalink]
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01 Apr 2012, 21:18
switching the tens and hundreds digits of a number does not affect its divisibility by 3. Remember the rule to find divisibility by 3? If the sum of all the digits is divisible by 3, then the number itself is divis. by 3.... so the sum won't change if you move the digits around. So since 400310 is equal to 90. We know that there should be around 90/3 numbers that are divisible by three... this is because every third number will be divisible by three. Note that this is an estimation... because if the series of numbers was 312 > 402, then we'd have one more. But we luckily don't need to figure that out, since the two closest answers to 30 are 22 and 90... so 30 it is!
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Re: How many threedigit integers between 310 and 400, exclusive [#permalink]
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11 Apr 2012, 04:27
rnemani wrote: How many threedigit integers between 310 and 400, exclusive, are divisible by 3 when the tens digit and the hundered digit are switched?
A. 3 B. 19 C. 22 D. 30 E. 90 Increment is 3 310  400 EXCLUSIVE, therefore find the closest number on each side that is divisible by 3: 312 and 399, this is inclusive > [(399312)/3]+1 = 29 + 1 = 30 Choice D How do I treat the information "when the tens digit and the hundered digit are switched?" though?



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Re: Factor problem [#permalink]
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17 Oct 2013, 11:17
x2suresh wrote: rnemani wrote: How many threedigit integers between 310 and 400, exclusive, are divisible by 3 when the tens digit and the hundered digit are switched? 1) 3 2) 19 3) 22 4) 30 5) 90 Totally, lost on the solutions and the approach, please help Answer is 30 No of digits between 310 and 400 (exclusive) = 4003101 = 89 No Intergers divisble by 3 =~ 89/3 ~ 30 Divsiblivity rule for 3 is Sum of all digits must be divisble by 3. It doesn't matter if hundred digit and tens digits are switched. e.g 372 is divisble by 3 (becae 3+7+2= 12 (divisble by 3)) switch digits 732 also divisble by 3 (becae 7+3+2= 12 (divisble by 3)) Divsiblivity rule for 3 is Sum of all digits must be divisble by 3. It doesn't matter if hundred digit and tens digits are switched.You have right. So it is enough to proceed like usual to find the numbers in a given range that are divisible by 3. So 399312/3 +1 =30.



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Re: How many threedigit integers between 310 and 400, exclusive [#permalink]
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25 Aug 2014, 01:20
divisibility by 3 requires the sum of the digits and hence the swapping of digits is not important. So 312 to 399 inclusive divided by three gives 30



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How many threedigit integers between 310 and 400, exclusive [#permalink]
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07 Sep 2015, 18:04
I think it would be A. As after switching the digits it should still be a number between 310 and 400. So we are left with 330,336 and 339



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How many threedigit integers between 310 and 400, exclusive [#permalink]
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07 Sep 2015, 21:21
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer. How many 3digits integers between 310 and 400, exclusive, are divisible by 3 when the tens digit and the hundreds digits are switched? a) 3 b) 19 c) 22 d) 30 e) 90 Let a 3digits integer be abc(=a*100 + b*10 +c*1). Since a*100 + b*10 +c*1 can be represented as a*99+b*9+(a+b+c), abc is divisible by 3 implies a+b+c is divisible by 3. So switching the tens digit and the hundreds digits does not affect on the divisibility by 3. We should find, therefore, the number of integers between 311 and 399, inclusive, which are divisible by 3. The first integer which is divisible by 3 is 312(3+2+1=6) and the last integer 399(3+9+9=21). So the number of integers which is divisible by 3 is (399312)/3 +1 = 29+1 =30. The answer is, therefore, D.
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Re: How many threedigit integers between 310 and 400, exclusive [#permalink]
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08 Sep 2015, 01:25
rakeshpedram wrote: I think it would be A. As after switching the digits it should still be a number between 310 and 400. So we are left with 330,336 and 339 Hi Rakesh, Although you have a nice and valid point here, but the question is not concerned about the state of the number after the swapping It simply asks How many integers between 310 and 400 are divisible by 3 when the tens digit and the hundred digit are switched After the switch the numbers may or may not be between 310 and 400.



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Re: How many threedigit integers between 310 and 400, exclusive [#permalink]
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09 Sep 2015, 01:26
TeamGMATIFY wrote: rakeshpedram wrote: I think it would be A. As after switching the digits it should still be a number between 310 and 400. So we are left with 330,336 and 339 Hi Rakesh, Although you have a nice and valid point here, but the question is not concerned about the state of the number after the swapping It simply asks How many integers between 310 and 400 are divisible by 3 when the tens digit and the hundred digit are switched After the switch the numbers may or may not be between 310 and 400. i'm confused, because the question says: "when the hundreds... reversed". does this not imply that we have to take care only about this special case? Wording is a bit confusing. Help appreciated



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Re: How many threedigit integers between 310 and 400, exclusive [#permalink]
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09 Sep 2015, 02:04
noTh1ng wrote: i'm confused, because the question says: "when the hundreds... reversed". does this not imply that we have to take care only about this special case? Wording is a bit confusing. Help appreciated We just need to find the numbers between 310 and 400 before the swap. After the swap the numbers may or may not be between 310 and 400. Moreover if we for once consider the numbers that will lie between 310 and 400, we have 330, 333, 336 and 339. We do not have 4 in any of the options. Hence, the question is not bothered about what happens to the number after the swap



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Re: How many threedigit integers between 310 and 400, exclusive [#permalink]
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03 Oct 2015, 04:15
Divisiblivity rule for 3 is sum of all digits must be divisible by 3 . So, switching the hundreds and tens digits will not change the sum . Hence , the new number will still be divisible . 312=3 x 104 ... 399=3 x 133 133104+1 =30
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Re: How many threedigit integers between 310 and 400, exclusive [#permalink]
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14 Mar 2016, 01:48



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Re: How many threedigit integers between 310 and 400, exclusive [#permalink]
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03 May 2016, 18:35
its simple 310 400
first number and the last number divisible by 3 is 312  399
so as per the formula to find the no of terms is last  first divide by 3 plus 1
so 399312/3 plus 1, so 87/3 plus 1 is 29 plus 1 equal 30
ans is 30



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Re: How many threedigit integers between 310 and 400, exclusive [#permalink]
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04 May 2016, 18:13



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Re: How many threedigit integers between 310 and 400, exclusive [#permalink]
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18 Jun 2016, 08:09
the wording of the question is not correct. we cannot assume whether the number should/not fall between given range after changing hundreds and tens. because option A makes it confusing.
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Re: How many threedigit integers between 310 and 400, exclusive [#permalink]
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25 Sep 2017, 03:43
rnemani wrote: How many threedigit integers between 310 and 400, exclusive, are divisible by 3 when the tens digit and the hundered digit are switched?
A. 3 B. 19 C. 22 D. 30 E. 90 The answer is D We know that the divisibility by 3 that the sum of the number must be divisible by 3 So we see that the interchanging of the digits of hundreds and tens will not impact out outcome The first number to be divisible is 312 and last number to be divisible is 399 so there are (399312)/3 +1=87/3+1=90/3=30 Numbers Hence D is our answer . I found this question to be little tricky
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