Divisible by 11 by 4 digit numbers using 5 and 0 only
Answers
Take the alternating sum of the digits in the number, read from left to right. If that is divisible by 11, so is the original number.
So, for instance, 2728 has alternating sum of digits 2-7+2-8 = -11. Since -11 is divisible by 11, so is 2728.
Similarly, for 31415, the alternating sum of digits is 3-1+4-1+5 = 10. This is not divisible by 11, so neither is 31415.
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Now the challenge is to find out the ordered pairs for such a combination in a range of 1000-9999.
By Simple Observation we can see the following ordered pairs:
{1,2,3,4}; {2,3,4,5}; {3,4,5,6}
{1,2,4,5}; {2,3,5,6}
{1,2,5,6};
{1,3,4,6};
Total Number of Ordered Pairs: 7
We take {1,2,3,4} as an example:
By the above description of test this ordered pair can be broken in 2 pairs:
{1,4} & {2,3}
Now the ones/hundredth digit can be filled in 2 ways only. { say _1_4 or _4_1}
Similarly the tenth/thousandth digit can be filled in 2 ways only. {say 2_3_ or _3_2}
Hence we will have 4 resultant numbers.
But the singular pairs of {1,4} & {2,3} can be altered in there respective positions which results in twice the number --> Hence 4*2 = 8 numbers.
Since there are 7 of such ordered pairs we will have 7*8= 56 answers.
ANSWER: 56
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For a number to be divisible by 11 we have the following test:
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Take the alternating sum of the digits in the number, read from left to right. If that is divisible by 11, so is the original number.
So, for instance, 2728 has alternating sum of digits 2-7+2-8 = -11. Since -11 is divisible by 11, so is 2728.
Similarly, for 31415, the alternating sum of digits is 3-1+4-1+5 = 10. This is not divisible by 11, so neither is 31415.
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Now the challenge is to find out the ordered pairs for such a combination in a range of 1000-9999.
By Simple Observation we can see the following ordered pairs:
{1,2,3,4}; {2,3,4,5}; {3,4,5,6}
{1,2,4,5}; {2,3,5,6}
{1,2,5,6};
{1,3,4,6};
Total Number of Ordered Pairs: 7
We take {1,2,3,4} as an example:
By the above description of test this ordered pair can be broken in 2 pairs:
{1,4} & {2,3}
Now the ones/hundredth digit can be filled in 2 ways only. { say _1_4 or _4_1}
Similarly the tenth/thousandth digit can be filled in 2 ways only. {say 2_3_ or _3_2}
Hence we will have 4 resultant numbers.
But the singular pairs of {1,4} & {2,3} can be altered in there respective positions which results in twice the number --> Hence 4*2 = 8 numbers.
Since there are 7 of such ordered pairs we will have 7*8= 56 answers.
ANSWER: 56
2.9k Views ·
The minimum sum of two digits can be 3 while maximum can be 11.
The divisibilty test of 11 says that sum of even digits - sum of odd digits should be a multiple of 11 or zero. Here since max sum is 11 and min sum is 3 the only possibilty left is zero. Hence we need to look for two pairs of digits that sum up same
For each number k between 3 and 11 we have two find two exlusive pair of numbers that sum up to k.
k = 3, {1,2}
k = 4, {1,3}
k = 5, {1,4}, {2,3}
k = 6, {1,5}, {2,4}
k = 7, {1,6}, {2,5}, {3,4}
k = 8, {2,6}, {3,5}
k = 9, {3,6}, {4,5}
k = 10, {4,6}
k = 11, {5,6}
For value of k where we have less than 2 pairs we cannot form the required number so their contribution is zero. For values of k with 2 pairs, we have to select both pairs and permute them accordingly. Let us say for k = 5, two pairs are {1,4} and {2,3}. Now is one occupies first place 4 will occupy third place. And the other pair will occupy second and fourth place. In other words one pair will occupy even places and one will acquire odd places.
So total no. of required numbers for k = 5 is:
no. of ways a pair can be selected for odd position*no. of ways this pair can be arranged in odd positions*no. of ways other pair can be arranged in even positions [math]= 2*2*2 = 8[/math]. Similar results are for k =6,8 and 9.
For k = 7, we first need to select two pairs out of three. So total no. of ways will be:
no. of ways in which two pairs can be selected out of three*no. of ways a pair can be selected for odd position*no. of ways this pair can be arranged in odd positions*no. of ways other pair can be arranged in even [math]= 3C2*2*2*2 = 24.[/math]
So, total such numbers will be:
4∗(2∗2∗2)+3C2∗(2∗2∗2)=32+24=56
1.8k Views ·
56 is the answer.
The number list was provided below
1243,1254,1265,1342,1364,1452,1463,1562,2134,2145,2156,2354
,2365,2431,2453,2541,2563,2651,3124,3146,3245,3256,3421,3465
,3542,3564,3641,3652,4125,4136,4213,4235,4312,4356,4521,4532
,4631,4653,5126,5214,5236,5324,5346,5412,5423,5621,5632,5643
,6215,6314,6325,6413,6435,6512,6523,6534
The description which I provided below may be stupid.
But I am sql developer . I have written the code in sql language
Hope this may be helpful for the sql developers.
Declare
a number(3) := 0;
begin
for i in 1234..9999 loop
if length(i) = length(translate(i,'1234567890','123456')) and mod (i,11) = 0 and (instr(i,1,1,2) + instr(i,2,1,2) + instr(i,3,1,2) + instr(i,4,1,2) + instr(i,5,1,2) + instr(i,6,1,2) = 0) then
a := a+1;
dbms_output.put_line(i);
end if;
end loop;
dbms_output.put_line('count is '||a);
end;