Study guides

Q: How do you find the consecutive integer of a square root number?

Write your answer...

Submit

Related questions

9,11,13

Try it! Find the square of some integers, until you either (a) find one that gives exactly 360 when squared, or (b) find two consecutive integers, one of which is smaller, the other larger, than 360. You can also take your calculator and find the square root of 360. If it is an integer (no decimals), that means it is a square number.

If you have two consecutive integers then one of them must be odd and the other must be even. The square of an odd integer must be odd, the square of an even integer must be even. The sum of an odd number and an even number must be odd. Thus, the sum of squares of any two consecutive numbers must be odd. Therefore, the question has no valid answer.

Find the square root of the number.Take the integer part of the answer.Square the integer part.Subtract this square from the original number.

Find 3 consecutive numbers where the product of the smaller two numbers is 19 less than the square of the largest number.

Pick a number. Add or subtract 1 to or from it.

Take any integer n and square it and you have a perfect square. Then you might want to know if a given number is a perfect square. Take the square root of a number and if it is a whole number, then the number is a perfect square.

6+7=13 and 6 and 7 are consecutive. 6 is the smaller integer so answer is 6

The integers are 12 and 14 (144+196=340)

Try squaring different integers (hint: in this case, the integers will be fairly small). If you find that the square of one integer is less than 27, and the square of the next integer is more than 27, you have your answer.

62 - 6 = 30 Therefore, the number is 6.

The numbers are 88 and 89 (88 x 89 = 7,832). This is very easy to find. The products you are looking for are consecutive and so they are very close to each other. Calculate the square root of the number, and it is easy to see that if there are two consecutive whole numbers that will be products of the number, they will be consecutive whole numbers very close to the square root. The square root of 7832 is about 88.5.

no solution. If you solve for x (where x is the first integer) the answer is a fraction, which is not an integer.

4,6,8

Square them both, find a non-square integer between those two results, and then take the square root of that number. In other words, find a non-square integer between 25 and 49, and since there is only one square number between them, 36, that should be easy; let's pick 42, and then take the square root of it. Ta da! âˆš42 is an irrational number between 5 and 7, its first 30 digits being 6.48074069840786023096596743608.

Just go over what you did to get your answer, and if you find a mistake, correct it.

31+32+33 = 96

It is: 5*6*7 = 120 and 5+6+7 = 18

Well, 47 49 51 53 are four consecutive odd numbers those total squared has for identical digits. 40000.... The square root of any number that is only four digits long all containing the same digit has a value that is not an integer.

As there are two consecutive integers then one must be an even number and the other an odd number. If the numbers are y and y + 1 then if y is even, y + 1 is odd and if y is odd then y + 1 is even.The product of an even integer and an odd integer is always even.The question therefore has no answer.121 = 112 but this is not what the question has asked.

four

The numbers are 9 and 10.

find the sum of 2 and 2

3,162.27766 is the square root of a smallest 8 digit number and thus the greatest integer having the greatest 7 digit square is 3,162. This is because the square of the next integer i. e. 3,163 is going to be an 8 digit number as discussed above. Thus, 3,162 is the required answer. 3162^2 = 9998244.

They are 14, 16 and 18.