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How many three-digit integers between 310 and 400? help...

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by egybs » Tue Jun 10, 2008 9:03 am
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!
Last edited by egybs on Tue Jun 10, 2008 9:06 am, edited 2 times in total.
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by vaivish » Tue Jun 10, 2008 9:03 am
hi...first of all for division of 3 rule changing digits would not make the differce as far as the sum of the digits is divisible by 3. Hence the change that the Q mentions will not have any difference.

Now the total nos divisible by 3 are : (400-310)/3= 30 . Here the inclusion of 310 does not come into the picture because it is not divisible by 3. If the first no. would have been say 309 then we would have to add +1 to 30. I hope this clears the doubt....
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by vittalgmat » Tue Jun 10, 2008 9:39 am
Hi,
Here is my logic.

Between 1 and 100 there are 33 integers which are divisible by 3. (ie.
100/3 ~= 33 ). This is same whether u start from 1 or 100 or 200 or 300.
So for numbers 310 to 400 both digits EXCLUSIVE (does not matter even if it was inclusive), there are 33 -3 = 30 numbers.

Here the switching of tenths and hundreds digit is a diversion. Coz the divisibility rule for 3 says "if the sum if the digits is divisible by 3 then then umber is divisible by 3".

Hope this helps.

thanks
-Vittal
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by antyagi » Tue Jun 10, 2008 9:50 am
Can this formula even used when there is condition that
" the tens digit and the hundreds digit are switched"
This would transform number into
310 -> 130
320 -> 230
330 -> 330
.
.
.
.
390 -> 930

Is the method of solution still valid ? I know the answer is correct
Thanks,
Ankur Tyagi
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by egybs » Tue Jun 10, 2008 12:35 pm
Yep! from my last post: "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."



If you have the digits x y and z

xyz is divisible by 3 if x+y+z is itself divisible by 3. So xzy, yzx, etc are all divisible by 3 if xyz is.

Try it out.

123 is divisible by three because 1+2+3 = 6 which is divisible by 3
321 is visible by three because 3+2+1 = 6 which is also divisible by three.

Hope this helps!







antyagi wrote:Can this formula even used when there is condition that
" the tens digit and the hundreds digit are switched"
This would transform number into
310 -> 130
320 -> 230
330 -> 330
.
.
.
.
390 -> 930

Is the method of solution still valid ? I know the answer is correct
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by llewellyn27 » Tue Jun 10, 2008 1:45 pm
Based on the formula for a series from a previous post

An = A1 + (n-1)*d

The 10th and units place are changed. So the resulting numbers would be 130 and 40 (Original numbers were 310 and 400)

So now we have to find the numbers between 40 and 130 that are divisible by 3

The last number before 130 that is divisible by 3 is 129 and the first number divisible by 3 is 42, d = 3

Using above equation:

129 = 42 + (n-1) * 3

Solve for n and find that n = 30

Please give me some feedback on this approach.
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by netigen » Tue Jun 10, 2008 2:07 pm
This is a poorly written question.

Whose tens and hundreds digits are switched?
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by harishamid » Tue Aug 24, 2010 12:26 pm
We can still use the formula, and do it in a different way.

an = a + (n-1)*d

Between 310 and 400 inclusive, 312 is the first number and 399 is the last number that are divisible by 3.

so above equation becomes:

399 = 312 + (n-1)*3
=> 29 = n-1
=> n = 30.
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by saurabhdhakad » Mon Jan 16, 2012 9:26 pm
I think answer could be 3 also.
As per the Q if the Hundreds and Tens place interchange, we should also be looking for the condition in the Q that no should be between 310 and 400 even after that.

Considering that, 333 : 336 : 339 are the only probable options.

I got the expected answer too.
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by LalaB » Tue Jan 17, 2012 12:12 am
310-400 inclusive
number is divisible by 3, if the sum of all its digits is divisible by 3. it means that switching the place of digits will not change the divisibility of numbers.

the 1st and last numbers divisible by 3 are 312 and 399 respectively.

so, (399-312)/3+1=30
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by ronnie1985 » Wed Jan 18, 2012 12:50 am
This calls for nos of the form 3XY such that X+Y = 3, 6, 9, 12, 15, 18 and x>=1
There are 28 such numbers
Also we need to count the numbers of the form 4XY such that X+Y =2, 5, 8. and X = 0
There are 3 such numbers. Answer should be 31 not 30.
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by ronnie1985 » Wed Jan 18, 2012 2:09 am
Sorry I made a mistake. The question asked the no of numbers divisible by 3 even if the tens and hundreds digit are swapped. Since the divisibility rule of 3 calls for sum of the digits ot there positions, the answer coincides with the answer of the question:find out the number of multiples of 3 in between 310 and 410 exclusive.

Answer = (408-312)/3+1 = 33
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by jayoptimist » Wed Jan 18, 2012 2:58 am
My Bit.

1. If the sum of the digits of an integer are divisible by 3, then the number is divisible by 3.
( This helps us resolve the problem of interchange of digits of 100's place and 10's place)
2. How many multiples of 3 are there between 310 and 400 ?
3*104 = 312
3*133 = 399

133-104+1 = 30.( Hope it is clear why I added +1)
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