How many 6 digit numbers can be formed from 1 to 6 which are divisible by 4?
We have received solutions from Emily and Ellie from Fitzharrys School, Victor from St. Paul's in Brazil, Jack from Springwood Academy, Rebecca from Loughborough High School, Komal from Alexandra School in India, Sam, Nathan and Dan from Netherthong Primary School, Anurag from Queen Elizabeth's Grammar School in Horncastle, Tom, Jake, Harry and Michael from Corsham Primary School, George from Onslow St. Audrey's School, Ali from Riccarton High School in Christchurch, New Zealand, and Luke from Cottenham in Cambridgeshire. Victor found that 12365 could be made a little longer: Yes, 123654 is divisible by 6, but they can't go any futher than that. He had a go at finding the longest number you can make that satisfies the rules of the game. He managed a seven digit number: 7836542 is my attempt, I didn't manage to use all of the digits. Jack found a six digit number that satisfied the rules: 147258 Emily & Ellie found a seven digit number: 9632581 It is possible to find a ten digit number that satisfies the rules. It is: 3816547290 since 3 is divisible by 1 Tom, Jake, Harry and Michael worked it out like this: We realised that the 0 had to go at the end (Units) for it to be a multiple of 10. Odd Even Odd Even 5 Even Odd Even Odd 0 We used the elimination method starting with 98 and went down to 92 and kept on going down. Eventually we found the answer: 3816547290. Ali worked it out like this: Let A B C D E F G H I J represent the 10 digits. The number itself is divisible by 10, so J must be 0. Since the number formed by the first 5 digits from the left is divisible by 5, E must be 5 or 0, but since we have used 0 already, E must be 5. Since AB, ABCD, ABCDEF and ABCDEFGH must be divisible by 2, 4, 6 and 8 respectively, they must all be even. Therefore B, D, F and H must be even numbers. That leaves 1, 3, 7 and 9 for A, C, G and I. Ali then started listing the possible combinations, and eliminating those that did not fit: AB could be any of the following: The number formed by the first 3 digits from the left is divisible by 3, so the digits must add up to a number that is divisible by 3. The possible values of ABC can now be identified: The number formed by the first 4 digits from the left is divisible by 4, so the number formed by the last two digits of the number must also be divisible by 4 (since all multiples of 100 are divisible by 4 we can ignore the digits in the hundreds and thousands column). The possible values of ABCD can now be identified: We know that the fifth digit must be 5, so the possible values of ABCDE can now be identified: The number formed by the first 6 digits from the left is divisible by 6, so it must be even, and the digits must add up to a number that is divisible by 3 (since it must be a multiple of 3). The possible values of ABCDEF can now be identified: The number formed by the first 7 digits from the left is divisible by 7, and the seventh digit must be odd. The possible values of ABCDEFG can now be identified: The number formed by the first 8 digits from the left is divisible by 8, so the number formed by the last three digits of the number must also be divisible by 8 (since all multiples of 1000 are divisible by 8 we can ignore the digits in the thousands, ten thousands and hundred thousands column). That leaves only one option for ABCDEFGH: Therefore ABCDEFGHIJ can only be: Here is Anurag's solution. Luke explained his reasoning like this: I knew that the last digit was zero because only numbers that have zero at the end are divisible by ten. xxxxxxxxx0 Numbers that are divisible by 5 end in 5 or 0, and 0 has already been used so the fifth digit must be 5. xxxx5xxxx0 Since an odd number is not divisible by an even number, the pattern must go odd, even, odd, even,.. The second digit is even and, since multiples of 100 are divisible by 4, the number formed by the first 4 digits from the left will be divisible by 4 if the number formed by the last two digits is divisible by 4, so, ignoring the first two digits, So these are our options: xxx25xxxx0 xxx65xxxx0 The sixth digit is even and, since multiples of 200 are divisible by 8, the number formed by the first 8 digits from the left will be divisible by 8 if the number formed by the last two digits is divisible by 8, so, ignoring the first six digits Since the fourth digit must be 2 or a 6 as well that means that the second and sixth digits must be 4 or 8. So these are our options: x4x258x6x0 x4x658x2x0 x8x254x6x0 x8x654x2x0 This is the bit my Dad helped with. The sum of the first 3 digits must be divisible by 3. Also the sum of the first 6 digits must be divisible by 3. Therefore the sum of the fourth, fifth and sixth digits must be divisible by 3. So these are our options now: x4x258x6x0 x8x654x2x0 The first 3 digits must add up to 3. This gives us the following options: 147258x6x0 741258x6x0 183654x2x0 381654x2x0 189654x2x0 981654x2x0 789654x2x0 987654x2x0 Because the sixth digit is even, and the eighth digit is either 2 or 6, The ninth digit can be anything because the sum of all the digits 1 to 9 is divisible by 9. This gives us the following options: 1472589630 7412589630 1836547290 3816547290 1896543270 1896547230 9816543270 9816547230 7896543210 9876543210 Finally, if we check that the numbers formed by the first 7 digits from the left are divisible by 7, we find that the only possible answer is 3816547290 Rebecca also found 3816547290. She sent us this explanation: We can immediately say that the last digit has to be 0 because the whole number has to be divisible by 10. The 5th digit: The 0 has already been used, excluding the chances of the 5th digit being anything other than 5, as the 5th digit has to be divisible by 5. Even and Odd: We can also figure out that the 2nd, 4th, 6th, and 8th digits have to be even because no multiple of 2, 4, 6 or 8 can be odd. This means that the 1st, 3rd, 5th, 7th and 9th digits must be odd! The 4th and 8th digits: Any hundred can be divided by 4 so we have to focus on the 3rd and 4th digits. In the 4 times table, after every odd ten comes a 2 or a 6. For example: 32, 76, 52, 96 etc? What next? After this, the quickest method is to use trial and error, in a structured manner, starting at either the smallest or largest possible number. Komal sent us this explanation, which demonstrates what a structured approach might look like: Let our number be - Since our 10-digit number should be divisible by 10 hence the digit J should be 0. Now, our 2-digit number should be divisible by 2 therefore B will be an even number. Similarly 4-digit, 6-digit and 8-digit numbers will also be even numbers. This implies that A, C, G and I can be the remaining options i.e. 1, 3, 7, 9 For digit B, we know that ABCD is a multiple of 4. The multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40... Since C is an odd number (1, 3, 7, or 9) and D is an even number, the number D cannot be 4, 8, 0 because these numbers require the previous number to be an even number for the number ABCD to be divisible by 4. Therefore D can only be 2 or 6. Similarly the 8-digit number is divisible by 8 and hence should be divisible by 4. Using the divisibility rule for 4 we can conclude that H can only be 2 or 6. This implies that B and F can only be 4 or 8 (the remaining two numbers). The number of options for the final number are - For a number to be divisible by 3, the sum of its digits should be a multiple of 3. Now ABC and ABCDEF are both multiples of 3. If B is 4 then the sum of A and C can be 2, 5, 8, 11, 14, 17, ... Therefore A and C can be (1,7), (7,1) How many 6Therefore, there are a total of 192 numbers which can be formed using the digits 1,2,3,4,5,6 without repetition such that the number is divisible by 4.
How many six digit number can be formed using the digits 1 to 6?Hence, the correct answer is 720.
How many 6Hence, the answer is 192. Was this answer helpful?
How many 6Hence, Answer is 5!
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