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1. Perfect Square: 

Nội dung chính

    What number should be multiplied to 1575 to make it a square number?What is the perfect square of 1575?What least number should 1575 be divided to get a perfect square number?What is the smallest number by which 1100 must be multiplied so that the product becomes a perfect square?

A natural number x is a perfect square if there exists a natural number y such that x=y2. In other words, a natural number x is a perfect square, if it is equal to the product of a number with itself.

2. Properties of Square Numbers:

(i) A number ending in 2, 3, 7, or 8 is never a perfect square.

(ii) The number of zeroes in
the end of a perfect square is never odd. So, a number ending in an odd number of zeroes is never a perfect square.

(iii) Squares of even numbers are always even.

(iv) Squares of odd numbers are always odd.

3. General Properties of Perfect Squares:

(i) For any natural number n, we have n2=
Sum of first n odd natural numbers

(i) The square of a natural number, other than 1, is either a multiple of 3 or exceeds a multiple of 3 by 1.

(iii) The square of a natural number, other than 1, is either a multiple of 4 or exceeds a multiple of 4 by
1.

(iv) There are no natural numbers p.. and q such that p2=2q2

4. Square roots:

The square root of a given natural number n is that natural number which when multiplied by itself gives n as the product and we denote the square root of n by n
. Thus, n=m⇔n=mét vuông.

5. Finding Square Roots:

(i) In order to find the square root of a perfect square, resolve it into prime factors; make pairs of similar factors and take the
product of prime factors, choosing one out of every pair.

(ii) For finding the square root of a decimal fraction, make the even number of decimal places by affixing a zero, if necessary; mark off periods and extract the square root; putting the decimal point in the square root as soon as the integral part is exhausted.

6. Properties of Square Roots:

For positive numbers a and b, we have

(i) ab=a×b

(ii) ab=ab

(iii) The number of digits in the square number is either double or 1 less than double of the number of digits in the given number.

(iv) If the number of digits of the square number is n, then the number of digits in the square root of the number =n2       when n is even  
 =n+12  when n is odd   

Solution : (i)By doing factorisation of `16562` we get,
`2times7times7times13times13`.
We can see that `2` is left unpaired therefore we will divide `16562` by `2`.
`=16562div2=8281`
`8281` is a perfect square of `91`.
Hence, `2` is the number by which `16562` must be divided so that the result is a perfect square.

(ii)By doing factorisation of `3698` we get,
`2times43times43`.
We
can see that `2` is left unpaired therefore we will divide `3698` by `2`.
`=3698div2=1849`
`1849` is a perfect square of `43`.
Hence, `2` is the number by which `3698` must be divided so that the result is a perfect square.

(iii)By doing factorisation of `5103` we get,
`3times3times3times3times3times3times7`.
We can see that `7` is left unpaired therefore we will divide `5103` by `7`.
`=5103div7=729`
`729` is a
perfect square of `27`.
Hence, `7` is the number by which `5103` must be divided so that the result is a perfect square.

(iv)By doing factorisation of `3174` we get,
`2times3times23times2323`.
We can see that `2` and `3` are left unpaired therefore we will divide `3174` by `6`.
`=3174div6=529`
`529` is a perfect square of `23`.
Hence, `6` is the number by which `3174` must be divided so that the result is a perfect
square.

(v)By doing factorisation of `1575` we get,
`3times3times5times5times7`.
We can see that `7` is left unpaired therefore we will divide `1575` by `7`.
`=1575div7=225`
`225` is a perfect square of `15`.
Hence, `7` is the number by which `1575` must be divided so that the result is a perfect square.

Hint $rm, 1575 = 15!cdot! 100! +! 75 = 25,(15!cdot! 4! +! 3) = 5^2cdot 3^2cdotcolor#C00 7:$ lacks only one $,color#C00??,$ to become square.

Remark $ $ Suppose, instead, that $rm:n:$ is not $,1575,$ but is a bigger integer that is difficult to factor completely, but we are given that it is not square. First, we can rule out $rm:k = 9,$ and $,25,,$ since multiplying $rm:n:$ by a square does
not change the squareness of $rm:n.:$ Since $,63 = 7cdot 3^2,:$ we infer that if $rm:63,n = 3^2!cdot! 7,n:$ is a square then so too is the smaller $rm:7n.:$ This leaves only the possibilities $rm: k = 7:$ or $rm:15 = 3!cdot! 5.:$ To determine which $rm:kn:$ is square, we need only determine the parity of the power of any one of the primes $,3,5,7,$ in the factorization of $rm:n.:$ For example, in your case, once we have determined that $rm:n =
5^2,j, 5nmid j,:$ then we know $rm:k:$ must have have an even power of $5$, which excludes $rm:k=3!cdot!5,:$ leaving $rm:k = 7:$ as the only possible solution.

What number should be multiplied to 1575 to make it a square number?

So to make it a perfect square, we need to multiply 1575 with 7.

What is the perfect square of 1575?

Hence, the least multiple of 1575 which is a perfect square = 1575 × 7 = 11025.

What least number should 1575 be divided to get a perfect square number?

Hence, `7` is the number by which `1575` must be divided so that the result is a perfect square.

What is the smallest number by which 1100 must be multiplied so that the product becomes a perfect square?

Complete step-by-step answer:
Hence, [11] is the smallest number which when multiplied to [1100] gives the number that is a perfect square.
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