1. How many zeros are there at the right-hand end of 100!? 2. If the product of
ID: 3146316 • Letter: 1
Question
1. How many zeros are there at the right-hand end of 100!? 2. If the product of two integers is 27 x 38 x 52 x 711 and their greatest common divisor is 23 x 34 x 5, what is their least common multiple? 3. Use Euclid's algorithm to find the greatest common factor of the following pairs of integers, a and b (a) 3745 and 1172 (b) 192876 and 192918 4. Look back at your calculations for the previous exercise. Use those to help decide whether there are pairs of integers x and y which satisfy the following equations. If so, use the method of back substitution or other method with those calculations to find an integer-pair solution x, y (a) 3745x +1172 y=5 (b) 192876 x +1929 18y=10 In those cases where you think that no pair can exist, give a clear reason why not.Explanation / Answer
1) The number of trailing (right hand end) zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:
n/5+n/52+n/53+...+n/5k, where k must be chosen such that 5k<n.
According to above 100! has 100/5+100/25=20+4=24 trailing zeros.
2) LCM * HCF = product of numbers
LCM* 23 *34 * 5 = 27 * 38 *52 *711
therefore, LCM = 27-3 * 38-4 * 52-1 * 711
LCM= 24 * 34 *5 * 711
3) The Euclidean Algorithm for finding GCD(A,B) is as follows:
a) GCD of 3745 and 1172
=GCD of (1172*3 + 229) and 1172
=GCD of 1172 and 229
=GCD of (229*5+27) and 229
=GCD of 229 and 27
=GCD of (27*8 + 13) and 27
=GCD of 27 and 13
=GCD of (13*2 + 1) and 13
=GCD of 13 and 1 = 1
b) GCD of 192876 and 192918
=GCD of 192918 and 192876
=GCD of (192876*1 + 42 ) and 192876
=GCD of 192876 and 42
=GCD of (42 * 4592 + 12 ) and 42
=GCD of 42 and 12
=GCD of (12*3 + 6) and 12
=GCD of 12 and 6
=GCD of 6*2 + 0 and 6
=GCD of 6 and 0 = 6
4) dont have any info about previous exercise.
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