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4 x (5 x 3 - 2 x 1 )

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Q: How do you make number 52 using numbers 1 2 3 4 5?
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Related questions

The sum of two numbers is -42 the first number minus the second number is 52 Find the numbers?

= The sum of two numbers is -42 the first number minus the second number is 52 Find the numbers? =


What two numbers equal 52?

There is only one number that equals 52. It is . . . . . (wait for it) . . . . . 52 .


What to numbers can multiply to make 52?

4 x 13 = 52


What are number that are divisible by 52?

Well, darling, numbers divisible by 52 are any multiples of 52. So, you've got 52, 104, 156, and so on. Just take 52 and keep adding 52 to it, and you've got yourself a whole bunch of numbers divisible by 52. Math made easy, honey!


How do you get the number 47 using the numbers 2235?

52 - (2+3) = 47 If you can't combine two of the numbers together, then ((2 / .2) * 5) - 3 = 47 as well.


What 2 numbers equal 52?

The only number that equals 52 is 52. 4 x 13 equals 52 and 26x2 equals 52 so whoever put fifty-two equals only itself is pretty dumb because their other ways to make fifty to like 104 divided by 2 equals 52 so multiply and divide 52 by certain numbers and you can make fifty-two if you freaking try


What number can in to 52?

The numbers that go into 52 and leave no remainders are 1, 2, 4, 13, 26 and 52.


The sum of two numbers is -63 the first number minus the second is -41 find the numbers?

-22


How many odd numbers in 52?

There are 26 odd numbers in the range of 1 to 52. This can be calculated by dividing the total number of integers by 2, as every second number is odd when counting from 1. Therefore, 52 divided by 2 equals 26 odd numbers in the range.


What two numbers multiply together to make 52?

1 and 52 is one possible answer.


The sum of 3 consecutive number is 153 what are the numbers?

The numbers are 50, 51, and 52.


How many whole numbers are there between 52 and 73?

Between 52 and 73 there are 53, 54, ..., 72 a total of 72 - 53 + 1 = 20 numbers. Including 52 and 73 there are 22 numbers.