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

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.

Answer: B

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

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

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

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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).

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.

The answer is B.

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

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)

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

A number is divisible by 11, if the sum of digits at odd places is equal to sum of those at even places.

Here, the digits are 1, 3, 4, 5, 6 and their sum is 19. Since the sum is 19, the difference cannot be greater than 19, hence 0 or 11.

0 as a multiple of 11. This requires sum at both the places to be equal or each to be 19/2, which is not an integer and hence not possible.

11 as the other multiple of 11. This requires sum at both the places to have a difference of 11. Thus, S+S+11=19 or S=4, that is the sums will be 4 and 15.
Only possibility for sum of 4 is 1+3. Therefore, the numbers at even places will be 1 and 3, with 1 at higher place value as we are looking for smallest number.
_ 1 _ 3 _. Thus 3 is at tens place.

Although, we have got the answer as B, let us find the number.
The remaining three places will get filled up by the remaining numbers in ascending order. Thus, 4, 6 and 5 will get filled up as 4, 5 and 6..
Number = 41536