How many three-digit integers between 310 and 400, exclusive

Answer is 30

No of digits between 310 and 400 (exclusive) = 400-310-1 = 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))

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?

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 310-340 and since the pattern repeats there should be 10 between 340-370 and 370-400.

hence, there are 30 numbers.

IMO 30

HERE IT GOES..

IF WE SEE ALL THE THREE-DIGIT 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

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 400-310 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!

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

-> [(399-312)/3]+1 = 29 + 1 = 30

Choice D

How do I treat the information "when the tens digit and the hundered digit are switched?" though?

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 399-312/3 +1 =30.

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

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

Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


How many 3-digits 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 3-digits 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 (399-312)/3 +1 = 29+1 =30. The answer is, therefore, D.

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

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

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

133-104+1 =30
_________________

The Order Does not matter in the divisibility by 3
Hence no of multiples of 3 between 310 and 400 inclusive is what the question is asking
Hence 30
i.e. D

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 399-312/3 plus 1, so 87/3 plus 1 is 29 plus 1 equal 30

ans is 30

Okay the switching of the digits in any integer does not effect the divisibility rule for 3 which requires sum of digits to be divisible by 3
Hence 30

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.

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 (399-312)/3 +1=87/3+1=90/3=30 Numbers
Hence D is our answer .
I found this question to be little tricky