Determine the smallest possible number p such that 10500p is a perfect square
Pls answer me
10500 = 2^2 3 5^3 7
so, to get all even powers, multiply by 3*5*7 = 105
10500 * 105 = ____
Express 10500 as a product of its prime factors
10500= 2×2×3×5×5×5×7
=2^2×3×5^3×7
since we need a perfect square, the numbers 3, 5^3 and 7 don't satisfy the condition.
the missing numbers are 3, 5 and 7
p=3×5×7
p=105
To determine the smallest possible number p such that 10500p is a perfect square, we need to find a number that satisfies the condition.
To start, we can write 10500p as the product of its prime factors. Prime factorizing 10500, we get:
10500 = 2^2 * 3 * 5^2 * 7
Now, for 10500p to be a perfect square, all the exponents of the prime factors in the factorization must be even.
For the prime factor 2, the exponent is already even (2^2).
For the prime factor 3, the exponent is also already even (3^0).
For the prime factor 5, the exponent is odd (5^1). Therefore, we need to increase the exponent of 5 to make it even. The smallest exponent for 5 that makes it even is 5^2.
For the prime factor 7, the exponent is also odd (7^1). So, we need to increase the exponent of 7 to make it even. The smallest exponent for 7 that makes it even is 7^2.
Therefore, the smallest possible number p that makes 10500p a perfect square is given by:
p = 5^2 * 7^2 = 25 * 49 = 1225.
So, the smallest possible number p is 1225.