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Math Expert V
Joined: 02 Sep 2009
Posts: 65184
A five-digit positive integer N has all digits different and contains  [#permalink]

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7 00:00

Difficulty:   85% (hard)

Question Stats: 49% (02:31) correct 51% (02:22) wrong based on 55 sessions

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Competition Mode Question

A five-digit positive integer N has all digits different and contains digits 1, 3, 4, 5, and 6 only. If N is the smallest possible number such that it is divisible by 11, then what is the tens digit of N.

A. 1
B. 3
C. 4
D. 5
E. 6

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GMAT Club Legend  V
Joined: 18 Aug 2017
Posts: 6433
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: A five-digit positive integer N has all digits different and contains  [#permalink]

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divisiblity rule of 11 ; sum of odd place of no = sum of even place of no
possible no ; smallest ; 14356 ; IMO d ; 5

A five-digit positive integer N has all digits different and contains digits 1, 3, 4, 5, and 6 only. If N is the smallest possible number such that it is divisible by 11, then what is the tens digit of N.

A. 1
B. 3
C. 4
D. 5
E. 6
Director  P
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Posts: 673
Location: Poland
GMAT 1: 570 Q41 V27
WE: Engineering (Consumer Electronics)
Re: A five-digit positive integer N has all digits different and contains  [#permalink]

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A five-digit positive integer N has all digits different and contains digits 1, 3, 4, 5, and 6 only. If N is the smallest possible number such that it is divisible by 11, then what is the tens digit of N.

A. 1
B. 3
C. 4
D. 5
E. 6

For a number to be divisible by 11 One's digit + Hundred's digit + Ten Thousand's digit - Ten's digit - Thousandth's digit should be divisible by 11.

Because the subtraction requires to be 2 digit number (11) or 0 ->
Try to maximize the addition of 3 numbers
0 is not possible since the maximum addition of 2 numbers is 11 and thus rest of the 3 numbers when added equals 8

4+5+6-1-3 = 11 is the combination

Since N has to be smallest -> N -> 41536 --> Therefore Ten's digit is 3 -> Answer - B
SVP  D
Joined: 24 Nov 2016
Posts: 1674
Location: United States
Re: A five-digit positive integer N has all digits different and contains  [#permalink]

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Quote:
A five-digit positive integer N has all digits different and contains digits 1, 3, 4, 5, and 6 only. If N is the smallest possible number such that it is divisible by 11, then what is the tens digit of N.

A. 1
B. 3
C. 4
D. 5
E. 6

N: combination{13456} divisible by 11;
A number is div by 11, when:
difference between alternating digits is div by 11;
in this case, we need to find a difference of 11 or 0.

13456: [1+4+6]-[3+5]=11-8=3=invalid
13465: [1+4+5]-[3+6]=10-9=1=invalid
41536: [4,5,6]-[1,3]=15-4=11=valid

Ans (B)
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Re: A five-digit positive integer N has all digits different and contains  [#permalink]

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1
N = abcde
N is divisible by 11 => (b+d) - (a+c+e) is divisible by 11
Because (1+3)-(4+5+6) is divisible by 11
=> The minimum value of N is 41536
=> Choice B
Sloan MIT School Moderator V
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Posts: 1295
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GMAT 1: 580 Q43 V27
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Re: A five-digit positive integer N has all digits different and contains  [#permalink]

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2
A five-digit positive integer N has all digits different and contains digits 1, 3, 4, 5, and 6 only. If N is the smallest possible number such that it is divisible by 11, then what is the tens digit of N.

A. 1
B. 3
C. 4
D. 5
E. 6

A number is divisible by 11 if difference between sum of digits at odd places and sum of digits in even places is either '0' or divisible be 11.

Here sum of all digits 1,3,4,5 and 6 is 19.
Hence sum of digits at odd places must be '15' and sum of digits at even places must be '4' so that it sums to '19' and differs by '11'.

Now 15 is possible for following combinations:
645
456
564
.....
.....
.....
More importantly digits 1 and 3 cant take odd places eventually to sum '15' as 4 can't take even place to sum '4'. Hence N must start with 4.

Also, the thousand digit(even place) must be smaller than tens digit so that N is smaller.
So, Tens place must be taken by 3.

Hence N = 41536

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Re: A five-digit positive integer N has all digits different and contains  [#permalink]

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1
We have to take the difference of alternative digits to check for the divisibility of N.

By trial and error method, the difference can not be be zero and only be +11

So, the set of odd digits and even digits that would satisfy the divisibility rule of 11 is
Odd —> {4, 5, 6} &
Even —> {1, 3}

The smallest number that can be formed using the above is 41536

Ten’s Digit = 3

Option B

Posted from my mobile device
Intern  B
Joined: 25 Jan 2020
Posts: 3
Re: A five-digit positive integer N has all digits different and contains  [#permalink]

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1
N is divisible by 11 only when the difference between the sum of (tens digit + thousands digit) and the sum of (unit digit + hundred digit + ten thousands digit) is divisible by 11. (found on the internet)

We have:
1. the sum of (tens digit + thousands digit) and the sum of (unit digit + hundred digit + ten thousands digit) = 1+3+4+5+6 =19
2. from the above sum, we can say that the difference between the 2 above sum need to be 0 or 11. But 19 is not divisible by 2, so the difference between the two sum need to be 11.

=> the bigger sum is: (11+19)/2 = 15 and the smaller sum is: (19-11)/2 = 4 = 1+3 => the smaller sum must be the sum of (tens digit + thousands digit).

Because the question require the smallest possible value of N => the tens digit must be 3.

=> Answer: B (this question is too evil to me).
CR Forum Moderator D
Joined: 18 May 2019
Posts: 811
Re: A five-digit positive integer N has all digits different and contains  [#permalink]

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1
From the divisibility rule for 11, either the sum of the even position digits of N must equal to the sum of the odd position digits of N or the difference between the sum of the even position digits and the sum of the odd position digits of N must be divisible by 11.
We know that N is formed from the digits 1, 3, 4, 5, and 6
No combination of the sum of at least two of these numbers (even position digits) is equal to the sum of the remaining three numbers (odd position digits).
However 4+5+6 = 15 (sum of odd position digits of N) and 1+3=4 (sum of even position digits of N)
15-4=11 and 11 is divisible by 11. So we are concerned about the even position digits since we are looking for the tens digit which will occupy an even position in a 5 digit number.
For the smallest possible value, 1 must occupy the first even position. Hence 3 must be the tens digit.

Director  D
Joined: 25 Jul 2018
Posts: 731
A five-digit positive integer N has all digits different and contains  [#permalink]

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2
Let’s say that N= ABCDE.
—> (A+ C+E) —(B+ D )= must be equal to zero or multiple of 11

Now, we have 1,3,4,5,6 for five-digit integer N
—> smallest number should be 41536 ( divisible by 11)
—> (4+5+6)—(1+3) = 15–4 = 11
Tens digit of N is 3

Posted from my mobile device

Originally posted by lacktutor on 05 Feb 2020, 23:12.
Last edited by lacktutor on 20 Apr 2020, 02:45, edited 1 time in total.
Intern  B
Joined: 13 Jan 2020
Posts: 18
Re: A five-digit positive integer N has all digits different and contains  [#permalink]

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By trial and error method, the difference can not be zero and only be +11

So, the set of odd digits and even digits that would satisfy the divisibility rule of 11 is
Odd are $${4, 5, 6}$$ and
Even are $${1, 3}$$

The smallest number that can be formed using the above is 41536

Ten’s Digit $$= 3$$

Option B
Target Test Prep Representative V
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Re: A five-digit positive integer N has all digits different and contains  [#permalink]

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2
Bunuel wrote:

Competition Mode Question

A five-digit positive integer N has all digits different and contains digits 1, 3, 4, 5, and 6 only. If N is the smallest possible number such that it is divisible by 11, then what is the tens digit of N.

A. 1
B. 3
C. 4
D. 5
E. 6

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The rule for divisibility by 11 is that the difference between the sum of the odd-numbered digits and the sum of the even-numbered digits must be divisible by 11. For a five-digit integer abcde, this means that (a + c + e) - (b + d) must be divisible by 11.

Notice that the sum of the digits of the number is 1 + 3 + 4 + 5 + 6 = 19. Since 19 is odd, the difference between the alternating sums cannot be 0. Thus, if the five-digit integer is to be divisible by 11, we must have the difference between the alternating sums equal 11 or -11.

Notice that 15 - 4 = 11 and 4 - 15 = -11 (and 15 + 4 = 19). Thus, we will look for two groups of numbers, sums of which are 15 and 4. The only way to obtain a sum of 4 using the given numbers is 1 + 3 = 4; thus the second and fourth digits of the numbers must be 1 and 3. The remaining three numbers are 4, 5 and 6. The smallest five-digit integer we can form using the above restrictions is 41,536 (notice that 41,536 is divisible by 11; we get 41536/11 = 3776), and the tens digit of 41,536 is 3.

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# A five-digit positive integer N has all digits different and contains   