0 1 2 3 4 5 6 7 8 9 using 2 of these numbers together how many are multiples of 7?

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There are 13 two-digit multiples of 7.
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How many 3 digit numbers can be formed from the digits 1 2 3 4 5 6 7 8 9 if each digit is used only once?

Multiply the number of possible starting numbers by the number ofpossible middle numbers by the number of possible end numbers toget your result. In example, The possible starting numbers are 1, 2, 3, 4, 5, 6, 7,8, and 9. Let's say I picked nine as the starting number. Sinceyour question states tha (MORE)

What number can be divided by 2 3 4 5 6 7 8 9 and always give you a remainder of 1?

If you multiply all those numbers together you will get a number ofwhich they are all a factor. Each of them will divide evenly into it with no remainder. Thatnumber is 362,880 So, if we add 1 to that, we will always get a remainder of 1 whenwe divide by those numbers. . 362,881 / 2 = 181,440 +1 (MORE)

How many 4 digits numbers can be formed using 0 1 2 3 4 5 6 7 8 9?

We figure out how many possibilities exist for each digit, and multiply them all together. Think of the digits as slots that need to be filled: _ _ _ _. The first digit can be anything but 0; 0345 is not a 4-digit number, for example. So there are 9 choices for the first digit. All the subsequent d (MORE)

What number between 1 to 2000 has the factors 1 2 3 4 5 6 7 8 9 and 10?

10=5*2 9= 3*3 8= 2*2*2 7= 7 6= 3*2 5= 5 4= 2*2 3= 3 2= 2 1= 1 We need a product that include all of the combinations in the right column. Starting from the top: 10 gives us: 5*2 9 gives us: 3*3 8 gives us: 2*2 (we already have the third 2) 7 gives us: 7 6 gives us: (MORE)

Quick puzzle why are the numbers in this order 8 5 4 1 7 6 3 2 0?

Answer: Alphabetical order. . Eight. Five. Four. One. Seven. Six. Three. Two. Zero. Look at the first letter of each number, they are in alphabetical order. E for eight, F for five and four, but i before O for one, also before S for seven and six, which is before T for Three and Two, Which (MORE)

Using any maths Symbols or Signs prove it... 0 0 0 equals 6 1 1 1 equals 6 2 2 2 equals 6 3 3 3 equals 6 4 4 4 equals 6 5 5 5 equals 6 6 6 6 equals 6 7 7 7 equals 6 8 8 8 equals 6 9 9 9 equals 6?

the first one is: (0!+0!+0!)!=6 Because 0!=1 0!+0!+0!=3 and 3!=6 Just use factorial (1+1+1)! = 6 3 Factorial = 6 2+2+2 = 6 So Simple (3*3)-3 = 6 Also Simple Sqrt(4) + Sqrt(4)+ Sqrt(4) = 6 Sqrt(4) = 2 So 2+2+2 =6 5+(5/5) = 6 So Simple 6+6-6 = 6 Its quite simple 7-(7/7) = 6 Cuberoot(8) + (MORE)

How many 4 numeral combinations can you make from 1 2 3 4 5 6 7 8 9?

If you can't repeat digits and the order of the digits matters, you can make (9 x 8 x 7 x 6) = 3,024 patterns. If you can't repeat digits and the order of the digits doesn't matter, you can make (9 x 8 x 7 x 6)/(4 x 3 x 2 x 1) = 126 unique groups. If you can repeat digits and the o (MORE)

How many three digit numbers can be formed using 1-2-3-4-5-6-7-8-9-0?

Assuming you don't want repetition: You have 9 options for the first position (any of the digits except 0), 9 options for the second digit (any digit, expect the digit you used in the first position), and 8 options for the third position (any digit, except the two you used before). Therefore you hav (MORE)

What number can be divided by 1 2 3 4 5 6 7 8 9 and 10 without fractional leftover?

1 is not a valid answer. 1/2 = 0.5, .5 is the fractional leftover the result can be the following : either 0, since u can divide 0 with anything to get 0 back OR the calculation is the following : to have 2,5 and 10, we need at least a multiple of 10 to have 3,6 we need at least 30 to h (MORE)

Are there any numbers not starting with 1 2 3 4 5 6 7 8 and 9?

Yes 0 is a number. In the base-10 system, there are only those numerals for real numbers (positive and negative). In other bases, for example base-12 or "duodecimal" numbers, letters are used to indicate the other numerals, A for 11 and B for 12. Hexadecimal (base-16) is widely used in computing.

How many even 3-digit numbers can be formed from digits 0 1 2 4 5 7 9 if each digit can be used only once?

To solve this question, two cases must be considered: Case 1 : The three-digit even number ends with 0. If the three-digit number ends with 0, then the number must be even. After the digit 0 is used, six digits remain. After any one of these six digits is chosen to be the first digit, five d (MORE)

How many 5 digit numbers from 1 0 6 2 9 and 7 have digits whose product is equal to 4?

If you are not allowed to repeat the digits then the answer is clearly 0. If you are allowed to repeat digits then the only way you can possibly reach a product of 4 is by using a combination of 3 "1's" and 2 "2's". The possible combinations are therefore: 22111 21211 21121 21112 12211 12121 12112 (MORE)

What number can be divided by 2 3 4 5 6 7 8 9 and always gives remainder of 1 2 3 4 5 6 7 8 respectively?

Actually, an infinite amount of numbers could follow this pattern. To figure it out, we can use algebra: Let x be the number to divide by 23456789. So if there is a remainder than the answer of this division would some number, let's call it y + (12345678/23456789) So we would have the equation: x (MORE)