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There are 28706 such combinations.

5456 of these comprise three 2-digit numbers,

19008 comprise two 2-digit numbers and two 1-digit numbers,

4158 comprise one 2-digit number and four 1-digit numbers and

84 comprise six 1-digit numbers.

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โˆ™ 2014-06-07 14:45:41
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Q: 6 digit combination from 1 to 42?
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Use four 6s to make 42?

( 6 / 6 ) * (( 62) + 6 ) = 42 1 * ( 36 + 6 ) = 42 1 * 42 = 42 42 = 42


What is the to the combination for a lock is a three digit numberthe digits are the prime factors of 42 listed from least to greatest what is the combination for the lock?

6 x 7 = 42 (it helps to know the multiplication tables), so the prime factors are 2 x 3 x 7, and the combination is 237


What is two sixths of 42?

two sixths of 42 is: 2/6 × 42 = 2/6 × 42/1 = (2×42)/(6×1) = (2×7×6)/6 = 2×7 = 14.


What number comes next in the sequence 1 2 6 42 1806?

Option 1The sequence is found by multiplying each term by the number one greater than itself. First, 1 * 2 = 2. Then, 2 * 3 = 6. Then, 6 * 7 = 42, and so on.n*(n+1) = x (where n equals the previous x)1*(1+1) = 22*(2+1) = 66*(6+1) = 4242*(42+1) = 18061806*(1806 +1) = 3,263,442.Another way to consider it is that each term is being squared then added plus the original amount, so the second term will be equal to the first term squared, then add the first term.We find the equation (n^2) + n , where n is the previous term.So the next term is again 3,263,442.Option 2 For this option, we take each number and multiply it by the subsequent prime number, beginning in the same values and then diverging from the first choice.First we take 1 and multiply it by the smallest prime greater than itself - 2, and get 2.The next prime after 2 is 3 which gives us 2 * 3 = 6.Six times the next prime, 7, gives us 42.The number 42 times its nearest greater prime of 43 gives us 1806.Finally, 1806 times the next prime number (which is 1811) gives us 3,270,666.Option 3 The third approach to this series is a long process but seems logical too.Lets start. The original series is:1, 2, 6, 42, 1806, ?The second digit can be derived by multiplying the first digit by 2. Then the third digit can be derived by multiplying the second digit by 3. Then the fourth digit is third digit times 7. The fifth is fourth digit multiplied by 43.So we had a second string of series of multipliers which consists: 2,3,7,43,?To get the fifth multiplier, a pattern can be seen from this series. The second digit is first digit is multiplied by 1 plus 1. The third digit is second digit multiplied by 2 plus 1. The fourth digit is third digit multiplied by 6 plus 1.This in turn has created the sequence 1, 2, 6 which in the sequential nature of these sequences seems to be created by multiplying by subsequent integers. So the next term would be 6 * 4 = 24.Going back to the multipliers we find that the fifth digit is 1033 (24 * 43).2, 3, 7, 43, 1033Again, the second digit is the first digit multiplied by 1 plus 1. The third digit is the second digit multiplied by 2 plus 1. The fourth digit is the third digit multiplied by 6 plus 1. The fifth digit is now the fourth digit multiplied by 24 plus 1.Now lets go back to the original series:1, 2, 6, 42, 1806, ?Now we see the sixth digit is the fifth digit multiplied by 1033 which gives us 1,865,598.* * * * *Yet another possibility is to fit a polynomial to the given sequence. There are infinitely many polynomials of order 5 or above but considerT(n) = (1667n^4 - 16554n^3 + 57685n^2 - 82158n + 39384)/24.For n = 1, 2, 3, 4 and 5 it gives the required 5 numbers and T(6) = 8661.In reality, short of reading the mind of the person who posed the question, there is no way of determining which of the infinitely many solutions is the "correct" one.3,263,44210,650,056,950,806 etcThe formula is X2 + XSimplest way is:n=n*(n+1).=== === === === === === other way: 1 2 12*1/2=6 .... 1 2 6 126*2/6=42 ..... 1 2 6 42 12642*6/42=1806 .... 1 2 6 42 1806 12642186*42/1806=2940042 .... 1 2 6 42 1806 2940042There are at least two possible answers:1. You could fit the polynomial function:tn = (1667n4 - 16554n3 + 57685n2 - 82158n + 39384)/24 for n = 1, 2, 3, ...Following the logic, the 6th term is 8661.2. One alternative ist1 = 1 and tn+1 = tn*(tn + 1) for n = 1, 2, 3, ...That definition gives the next term as 3,263,442.There are many other solutions.1806*1807=32634423,263,442 comes after 1806. The series is formed by a number multiplying itself increased by 1. Here is the equation: X(X+1) Y1(1+1) 22(2+1) 66(6+1) 4242(42+1) 18061806(1806+1) 3,263,442In simpler numbers,1 X 2 22 X 3 66 X 7 4242 X 43 18061806 X 1807 3,263,4423263442. The progression is n(n+1).1 2 6 42 1'806 3'263'442 10'650'056'960'806 1st is 12nd is 1st*(1st+1) -> 1*23rd is 2nd*(2nd+1) -> 2*34th is 3rd*(3rd+1) -> 6*75th is 4th*(4th+1) -> 42*43...and so on32634423263442


What is one seventh of 42?

1/7 x 42 = 6

Related questions

6 digit number combination from 1 to 42?

3776965920


What 2 digit number x 1 digit number makes 378?

42 x 9 or 63 x 6


What 2 digit number times 1 digit number equals 378?

63 × 6 54 × 7 42 × 9


Use four 6s to make 42?

( 6 / 6 ) * (( 62) + 6 ) = 42 1 * ( 36 + 6 ) = 42 1 * 42 = 42 42 = 42


How do you use four 6's to make 42?

( 6 / 6 ) * (( 6^2) + 6 ) = 42 1 * ( 36 + 6 ) = 42 1 * 42 = 42 42 = 42


What is the to the combination for a lock is a three digit numberthe digits are the prime factors of 42 listed from least to greatest what is the combination for the lock?

6 x 7 = 42 (it helps to know the multiplication tables), so the prime factors are 2 x 3 x 7, and the combination is 237


How do you get the sum of the digit of a two digit number is 6?

33 is a two-digit number whose digits add up to 6. 60, 51, 42, 24, 15


What 2 digit number divisible by 4 6 7?

42, 84


What is 6 over 7 plus 1 over 6?

6/7 + 1/6 = 36/42 + 7/42 = 43/42 = 11/42


What is two sixths of 42?

two sixths of 42 is: 2/6 × 42 = 2/6 × 42/1 = (2×42)/(6×1) = (2×7×6)/6 = 2×7 = 14.


What is 6 divided 42?

0.1429


What number comes next in the sequence 1 2 6 42 1806?

Option 1The sequence is found by multiplying each term by the number one greater than itself. First, 1 * 2 = 2. Then, 2 * 3 = 6. Then, 6 * 7 = 42, and so on.n*(n+1) = x (where n equals the previous x)1*(1+1) = 22*(2+1) = 66*(6+1) = 4242*(42+1) = 18061806*(1806 +1) = 3,263,442.Another way to consider it is that each term is being squared then added plus the original amount, so the second term will be equal to the first term squared, then add the first term.We find the equation (n^2) + n , where n is the previous term.So the next term is again 3,263,442.Option 2 For this option, we take each number and multiply it by the subsequent prime number, beginning in the same values and then diverging from the first choice.First we take 1 and multiply it by the smallest prime greater than itself - 2, and get 2.The next prime after 2 is 3 which gives us 2 * 3 = 6.Six times the next prime, 7, gives us 42.The number 42 times its nearest greater prime of 43 gives us 1806.Finally, 1806 times the next prime number (which is 1811) gives us 3,270,666.Option 3 The third approach to this series is a long process but seems logical too.Lets start. The original series is:1, 2, 6, 42, 1806, ?The second digit can be derived by multiplying the first digit by 2. Then the third digit can be derived by multiplying the second digit by 3. Then the fourth digit is third digit times 7. The fifth is fourth digit multiplied by 43.So we had a second string of series of multipliers which consists: 2,3,7,43,?To get the fifth multiplier, a pattern can be seen from this series. The second digit is first digit is multiplied by 1 plus 1. The third digit is second digit multiplied by 2 plus 1. The fourth digit is third digit multiplied by 6 plus 1.This in turn has created the sequence 1, 2, 6 which in the sequential nature of these sequences seems to be created by multiplying by subsequent integers. So the next term would be 6 * 4 = 24.Going back to the multipliers we find that the fifth digit is 1033 (24 * 43).2, 3, 7, 43, 1033Again, the second digit is the first digit multiplied by 1 plus 1. The third digit is the second digit multiplied by 2 plus 1. The fourth digit is the third digit multiplied by 6 plus 1. The fifth digit is now the fourth digit multiplied by 24 plus 1.Now lets go back to the original series:1, 2, 6, 42, 1806, ?Now we see the sixth digit is the fifth digit multiplied by 1033 which gives us 1,865,598.* * * * *Yet another possibility is to fit a polynomial to the given sequence. There are infinitely many polynomials of order 5 or above but considerT(n) = (1667n^4 - 16554n^3 + 57685n^2 - 82158n + 39384)/24.For n = 1, 2, 3, 4 and 5 it gives the required 5 numbers and T(6) = 8661.In reality, short of reading the mind of the person who posed the question, there is no way of determining which of the infinitely many solutions is the "correct" one.3,263,44210,650,056,950,806 etcThe formula is X2 + XSimplest way is:n=n*(n+1).=== === === === === === other way: 1 2 12*1/2=6 .... 1 2 6 126*2/6=42 ..... 1 2 6 42 12642*6/42=1806 .... 1 2 6 42 1806 12642186*42/1806=2940042 .... 1 2 6 42 1806 2940042There are at least two possible answers:1. You could fit the polynomial function:tn = (1667n4 - 16554n3 + 57685n2 - 82158n + 39384)/24 for n = 1, 2, 3, ...Following the logic, the 6th term is 8661.2. One alternative ist1 = 1 and tn+1 = tn*(tn + 1) for n = 1, 2, 3, ...That definition gives the next term as 3,263,442.There are many other solutions.1806*1807=32634423,263,442 comes after 1806. The series is formed by a number multiplying itself increased by 1. Here is the equation: X(X+1) Y1(1+1) 22(2+1) 66(6+1) 4242(42+1) 18061806(1806+1) 3,263,442In simpler numbers,1 X 2 22 X 3 66 X 7 4242 X 43 18061806 X 1807 3,263,4423263442. The progression is n(n+1).1 2 6 42 1'806 3'263'442 10'650'056'960'806 1st is 12nd is 1st*(1st+1) -> 1*23rd is 2nd*(2nd+1) -> 2*34th is 3rd*(3rd+1) -> 6*75th is 4th*(4th+1) -> 42*43...and so on32634423263442

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