Squares of all integers are known as
perfect squares. In this lesson, we will discuss a very
interesting Mathematical shortcut: How to check whether a number is a perfect square
or not. There
are some properties of perfect squares which can be used to test if a number is
a perfect square or not. They can definitely say if it is not the square. (i.e.
Converse is not necessarily true).
All perfect squares end in 1, 4, 5, 6, 9 or
00 (i.e. Even number of zeros). Therefore, a number that ends in 2, 3, 7 or 8 is not a perfect square.
For all the numbers ending in 1, 4, 5, 6, &
9 and for numbers ending in even zeros, then remove the zeros at the end of the
number and apply following tests:
 Digital roots are 1, 4, 7 or 9. No number can be a perfect square unless its digital root is 1, 4, 7, or 9. You might already be familiar with computing digital roots. (To find digital root of a number, add all its digits. If this sum is more than 9, add the digits of this sum. The single digit obtained at the end is the digital root of the number.)
 If unit digit ends in 5, ten’s digit is always 2.
 If it ends in 6, ten’s digit is always odd (1, 3, 5, 7, and 9) otherwise it is always even. That is if it ends in 1, 4, and 9 the ten’s digit is always even (2, 4, 6, 8, 0).
 If a number is divisible by 4, its square leaves a remainder 0 when divided by 8.
 Square of even number not divisible by 4 leaves remainder 4 while square of an odd number always leaves remainder 1 when divided by 8.
 Total numbers of prime factors of a perfect square are always odd.
4539
ends
in 9, digit sum is 3. Therefore, 4539 is not a perfect square
5776
ends
in 6, digit sum is 7. Therefore, 5776 may be a perfect square
Step 1: A perfect square never ends in
2, 3, 7 or 8
This is the first observation you will
make to check if the number is a perfect square or not. For example, consider
the example 15623.
15623
By just noticing the number itself, we
can conclude that 15623 cannot be a perfect square. We do not have to go to Step
2.
Step 2: Obtain the digital root of the number:
How does the digital root of a number
would help in determining if a number is a perfect square or not. It turns out;
a perfect square will always have a digital root of 0, 1, 4 or 7.
Take the number 15626 for example. This
number ends in digits 6. So it satisfies Step 1. But still we cannot conclude,
this number as a perfect square.
Let’s take the digital root of this
number.
So, the digital root of this number is
2. A perfect square will never have a digital root 2. Hence, we can conclude
15626 is not a perfect square.
Now, there is a rider for this shortcut though, even if both Steps are satisfied, that does not guarantee that the number is a perfect square.
Let us take up an example here.
Consider the number 623461,
which is not a perfect square.
Notice that the unit digit is 1. This number could be a perfect square. Let us take the digital root.
The digital root of 623461 is 4. So it satisfies both Step 1 and 2. Still we cannot conclude that 623461 is a perfect square though.
However, this shortcut comes in really handy to eliminate obvious choices which are not a perfect square to solve competitive examination where you need to find the perfect squares.
Is 14798678562 a perfect square? Is 15763530163289 a perfect square?
Examine
both the units digits and the digital roots of perfect squares to help
determine whether or not a given number is a perfect square.
As
we know a perfect square can only end in a 0, 1, 4, 5, 6, or 9; this should
allow us to determine whether the first of our numbers is a perfect square.
However, it isn't sufficient to draw a conclusion about the second number.
Again
as we know that if a perfect square ends in 9; it’s tens digit is always even. Alas,
even if we do this, it won't rule out numbers ending in 89, because '...89' is
a possible square.
However,
as we know no number can be a perfect square unless its digital root is 1, 4,
7, or 9; so, find the digital root of our second number. It’s 5. As 5 isn't in
this list, then the number is definitely not a perfect square.
So, we can conclude, a number cannot be an
exact or perfect square if

it ends
in 2, 3,7 or 8

it terminates
in an odd number of zeros

its last
digit is 6 but its penultimate (tens) digit is even

its last
digit is not 6 but its penultimate (tens) digit is odd

its last digit is 5 but its penultimate (tens) digit is other
than 2

its last
2 digits are not divisible by 4 if it is even number
nice
ReplyDeleteI read in the Guardian recently (Chris Maslanka's puzzles) that no perfect square can leave a remainder of 3 or 5 on division by 7. No explanation or justification was given. Do you know of any?
ReplyDeleteLook at all possible remainders upon division by 7 (so 0 through 6) and see what the remainder is when you divide 0^2, 1^2, ..., 6^2 by 7. The possible remainders are 0, 1, 4, 2, 2, 4, and 1. So no perfect square gives a remainder of 3, 5, or 6 upon division by 7.
Delete679/7=97
DeleteReminder is 0 but it is not perfect square number
It is not perfect square because, the last digit is 9(odd) and the penultimate digit is 7 which is also odd. The penultimate digit has to be an even number, to make the number perfect square.
DeleteWhat about 99976 which give previous digit to be 7odd ,on division it gives remainder to be 1
Deletetotal number of prime factors is always even for a perfect square...total number of factors is odd...
ReplyDeleteex:36=2^2 * 3^2
total no. of factors=(2+1)*(2+1)=6
total number of distinct prime factors=2(2,3)
correct me if i am wrong.thank you
just a small typo (2+1)*(2+1)=9 as u said it always has to be odd hence its true
DeleteThanks
ReplyDeleteAccording to these rules are 61 and 76 perfect squares
ReplyDelete514 is a perfect square??...
ReplyDeleteIt's digital root is 1
Nope, it's last 2 digits are not divisible by 4
Deletenice...
ReplyDelete81 = 9^2?
ReplyDeletedigital root is 9 so it fails on step 2
Is the digital root the iterative sum of the digits until there is only one digit or is it the remainder when the number is divided by 9?
Digital root is the sum of all digits to get one digit. 81 does not fail step two as its digital root is 9. Numbers with digital roots 2,3,7,8 are not perfect squares..
DeleteS siva:
ReplyDelete514%7=3, so its not a perfect square.
what about the no like 10,1000. these numbers are not perfect squares. This method fails at step II.
ReplyDeleteYou might not need to go to step 2 here. Step 1 rules out the possibility of perfect square as it ends with odd number of zeros.
Deleteis X64 perfect square??
ReplyDeleteWhat about 2356
ReplyDeleteIt satisfies unit digit , digital root, unit digit 6 and tens place odd no, last two digit divisible by 4 and even no.
So where does ur method stand ?
100856 IS PERFECT SQUARE OR NOT ?
ReplyDelete1. ENDS WITH 6
2. DIGITAL ROOT = 2
3. UNIT DIGIT =6 , AND TENS DIGIT IS ODD,
4. LAST TWO DIGIT 56, IS DIVISIBLE BY 4.
ANYONE CAN EXPLAIN ?
It's clearly stated that there are "riders" over and above these steps that these methods can only be used as a means of finding which number "cannot" be a perfect square ! Also, that even if one condition fails, the number fails to be a perfect square .. the conclusion sums it up perfectly 
ReplyDelete" A number cannot be a perfect square if ....... "
15763530163289 is perfect square or not?
ReplyDeleteIs 250'and 5050 a perfect square
ReplyDeleteNo because ends with odd number of zeroes
Delete1146600 it ends on 00, digital root is 9 but still it is not perfect square
ReplyDeleteafter removing 00 you get 6 as unit digit but tens digit(6) is even
Deleteso it is not a perfect square.
Is there a method for binary numbers?
ReplyDelete9981 is perfect square or not....can any one explain this correctly
ReplyDeleteUsing this method we can only determine if a number is not a perfect square. That to only to an extent. You still need factorization to come to a final conclusion.
ReplyDeletewhat about 4356
ReplyDeleteAs per above explanation
step 1 satisfied
step 2 not satisfied. Digital root is 9.
But still it is a perfect square of 66
I dont think this concept will work always.
It satisfies. Digital root should be 1, 4, 7, or 9.
Delete1021
ReplyDeleteA few numbers can't be checked by using this method.
ReplyDeleteAccording to your rules 99976 should be a perfect square but it is not a perfect square number
ReplyDeleteSome are saying this rule fails and all... common fellas..
DeleteThis rule helps to find if a number is a imperfect square...
if a number let say 2356 satisfies all above rules, still we cant guarantee that the no is a perfect square. it only helps to channelise imperfect squares and not to tell that it is a perfect square
This trick hasn't any sense... Not satisfying all conditions
Delete184 is not a perfect square whil it's digital root is 4.
ReplyDeleteCan anybody tell me why???
If a perfect square is divisible by 4, but not 8, its remainder from 32 is always 4
ReplyDeletei want reasons to show that 13,32,18,4487 are non perfect squares
ReplyDeleteSubtract 1 from given number and see that obtained number divisible by 3 or sum of digits of obtained number is 8, then the given number is perfect square.
ReplyDelete1024 is a perfect square (of 32)
ReplyDeleteHowever digital root is 7. Explain?
What about the number 945729 ? Its digital root is 9 but its not a perfect square.How?
ReplyDeletewe have two conditions.
Delete1. if number ends with 2,3,7,8 its not an perfect square.
2. if digtal root is 1,7,4,9 the number may be a perfect square.
means if digital root is other than 1,7,4,9 we can say for sure that the number is not perfect square but the converse is not true
Square of even number not divisible by 4 leaves remainder 4 while square of an odd number always leaves remainder 1 when divided by 8?
ReplyDeletewhat do you mean. plz explain with example
Square of even number not divisible by 4 leaves remainder 4 while square of an odd number always leaves remainder 1 when divided by 8?
ReplyDeletewhat do you mean. plz explain with example
Please explain this :
ReplyDeleteIf a number is divisible by 4, its square leaves a remainder 0 when divided by 8.
Square of even number not divisible by 4 leaves remainder 4 while square of an odd number always leaves remainder 1 when divided by 8.
Thank u sooooo much i am a student of 7th when I was in 6th there was nothing hard to search on internet but now suddenly I have a very hard course of maths subject and first time I am understood at first time and without any confusion now I'm the fan of this "Burning math"..... Again thanks I am vaishnavi. S. Singh
ReplyDelete