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how to know whether a six or more diit no. is a perfect square or not? plz also give me the divisibility test of all nos.?

how to know whether a six or more diit no. is a perfect square or not?


plz also give me the divisibility test of all nos.?

Grade:12

1 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
7 years ago
Hello student,
Please find the answer to your question below
All perfect squares end in 1, 4, 5, 6, 9 or 00 (i.e. Even number of zeros). Therefore, a number thatends in 2, 3, 7 or 8 is not aperfect square.
Digital rootsare 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 withcomputing 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.
A number isdivisible by 2if its last digit is also (i.e. 0,2,4,6 or 8).
A number isdivisible by 3if the sum of its digits is also. Example: 534: 5+3+4=12 and 1+2=3 so 534 is divisible by 3.
A number isdivisible by 5if the last digit is 5 or 0.
Test for divisibility by 7. Double the last digit and subtract it from the remaining leading truncated number. If the result is divisible by 7, then so was the original number. Apply this rule over and over again as necessary. Example: 826. Twice 6 is 12. So take 12 from the truncated 82. Now 82-12=70. This is divisible by 7, so 826 is divisible by 7 also.
There are similar rules for the remaining primes under 50, i.e. 11,13, 17,19,23,29,31,37,41,43 and 47.
Test for divisibility by 11. Subtract the last digit from the remaining leading truncated number. If the result is divisible by 11, then so was the first number. Apply this rule over and over again as necessary.
Example: 19151--> 1915-1 =1914 -->191-4=187 -->18-7=11, so yes, 19151 is divisible by 11.
Test for divisibility by 13. Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. Apply this rule over and over again as necessary.
Example: 50661-->5066+4=5070-->507+0=507-->50+28=78 and 78 is 6*13, so 50661 is divisible by 13.
Test for divisibility by 17. Subtract five times the last digit from the remaining leading truncated number. If the result is divisible by 17, then so was the first number. Apply this rule over and over again as necessary.
Example: 3978-->397-5*8=357-->35-5*7=0. So 3978 is divisible by 17.
Test for divisibility by 19. Add two times the last digit to the remaining leading truncated number. If the result is divisible by 19, then so was the first number. Apply this rule over and over again as necessary.
EG: 101156-->10115+2*6=10127-->1012+2*7=1026-->102+2*6=114 and 114=6*19, so 101156 is divisible by 19.

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