all the numbers,are divisible by 4, 3times 4times 1,2 times 4 times 1,and2times3 times 4,one must alway's try thinking outside the box,as in this question,make 15 out of six ones?,well there are two ones in the number 11,1 in 12,1,in 13,1 in 14,1 in 15,add up the ones it makes 15
What is 4 divided by 1,234
4(2+3)-1=19
How about: 2*(1+3+4) = 16
The biggest number that can be formed with just those four digits once each, is 8641 - however - no number made from those digits can be divided by 9 because in order for a number to be exactly divisible by nine - the sum of the digits must also divide exactly by nine. The sum of the digits 6, 4 8 & 1 is 19 !Another Answer:-The above answer is absolutely correct.A possible solution is 61+8/4 = 63 and 7*9 = 63
To get the number 16 using only the digits 1, 2, 3, and 4, you can use the following mathematical expression: (4 x 3) + (2 x 1) = 12 + 4 = 16. This equation involves multiplication and addition, using each of the given digits exactly once.
There are none.Any number made by using the digits 0 to 9 once is divisible by 9 and so not a prime.There are none.Any number made by using the digits 0 to 9 once is divisible by 9 and so not a prime.There are none.Any number made by using the digits 0 to 9 once is divisible by 9 and so not a prime.There are none.Any number made by using the digits 0 to 9 once is divisible by 9 and so not a prime.
972.
total 6 they are 3412,4312,1324,3124,1432,4132
A delectable number has nine digits, using the numbers 1-9 once in each digit. The first digit of a delectable number must be divisible by one. The first and second digits must be divisible by two, the first through third must be divisible by three, etc. There has only been one delectable number discovered: 381654729.
The difference is 360.
What is 4 divided by 1,234
4(2+3)-1=19
The answer is 34(1)^2.
How about: 2*(1+3+4) = 16
18 = (4 × 3) + (2 × 1)
17 = (4+2)*3 - 1
Every palindrome with an even number of digits is divisible by 11. The easiest way to see this is to recall the divisibility rule by 11: if a number X is written as ABCDEFG... (here A,B,C, ... are digits), then it's divisible by 11 if and only if the sum A-B+C-D+E-F+G-... is divisible by 11. In a palindrome with an even number of digits, each digit will appear in an odd position and in an even position, so when we calculate this sum, it will be added once and subtracted once, canceling. Since all the digits cancel, the sum A-B+C-D+... will be 0, which is divisible by 11. So the original number ABCD....DCBA was also divisible by 11.