You cannot use exponents for this problem. 7x4x2
yes you can it is 7x23
Since '112' is an even number, you divide by '2' , the smallest prime number. 2)112 2)56 2)28 2)14 7)7 = 1 NB Reduce by using prime number denominators until you reach '1'. Hence 112 = 2x2x2x2x7 = 2^(4) X 7^(1)
23 x 7 = 56
Prime factorization of 7 = 7 Prime factorization of 56 = 2*2*2*7 Common number(factor in factorization) or simply H.C.F.= 7 L.C.M. = H.C.F. * Product of other numbers = 7*2*2*2 = 56 In this case 7 is an exact divisor of 56, therefore L.C.M. = 56
To find the Least Common Multiple (LCM) of 98, 56, and 28, we first need to find the prime factorization of each number. The prime factorization of 98 is 2 * 7^2, the prime factorization of 56 is 2^3 * 7, and the prime factorization of 28 is 2^2 * 7. To find the LCM, we take the highest power of each prime factor that appears in any of the numbers. Therefore, the LCM of 98, 56, and 28 is 2^3 * 7^2, which equals 392.
Prime Factorization of 24 and 56Prime factorization of 24 is:2 X 122 X 2 X 62 X 2 X 2 X 3The prime factorization of 56 is:2 X 282 X 2 X 142 X 2 X 2 X 7
23 x 71 = 56
23 x 7 = 56
56 prime factorization 2*2*2*7 exponents: 23*7 * = multiply
28 x 56 = 4,000,000
56 = 2 x 2 x 2 x 7 = 23 x 7
Prime Factorization of 58 Using ExponentsThe prime factorizatio of 58 is:2 X 29Exponents are not needed because only one 2 and one 29 are the prime factors of 58. If exponents were used, it would be 21 X 291.
Since '112' is an even number, you divide by '2' , the smallest prime number. 2)112 2)56 2)28 2)14 7)7 = 1 NB Reduce by using prime number denominators until you reach '1'. Hence 112 = 2x2x2x2x7 = 2^(4) X 7^(1)
The prime factorization of 56 is:7 x 23 = 56
56 = 2 * 2 * 2 * 7 = 23 * 7
It is: 4
whats continuous divisions of 56
4000000 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 5 * 5 * 5 * 5 * 5 * 5 = 28 * 56