Question 6973
a. 45,36,24 
{{{GCD = 3}}} 
{{{LCM= 360}}}


b. 42,96,104,18 
{{{GCD = 2}}} 
{{{LCM = 26,208}}} 


In order to find the GCD, let's look at their prime factorization:

PROBLEM #1:  GCD
45: {{{2^0}}}  x  {{{3^2}}}  x  {{{5^1}}}
36: {{{2^2}}}  x  {{{3^2}}}  x  {{{5^0}}}
24: {{{2^3}}}  x  {{{3^1}}}  x  {{{5^0}}}


Next, find the MINIMUM exponent of each base:
for 2, the minimum exponent is 0, written as {{{2^0}}} 
for 3, the minimum exponent is 1, written as {{{3^1}}}
for 5 the minimum exponent is 0, written as {{{5^0}}}


Multiply the minimum exponent of each base and you get: 
{{{2^0}}}  x  {{{3^1}}}  x  {{{5^0}}}  = 3

******THE GCD FOR 45,36,24 = 3*****




PROBLEM #2:  GCD
42:    {{{2^1}}}  x  {{{3^1}}}  x  {{{7^1}}}  x  {{{13^0}}}
96:    {{{2^5}}}  x  {{{3^1}}}  x  {{{7^0}}}  x  {{{13^0}}}
104:  {{{2^3}}}  x  {{{3^0}}}  x  {{{7^0}}}  x  {{{13^1}}}
18:    {{{2^1}}}  x  {{{3^2}}}  x  {{{7^0}}}  x  {{{13^0}}}


Next, find the MINIMUM exponent of each base:
for 2, the minimum exponent is 1, written as {{{2^1}}} 
for 3, the minimum exponent is 0, written as {{{3^0}}}
for 7 the minimum exponent is 0, written as  {{{7^0}}}
for 13 the minimum exponent is 0, written as {{{13^0}}}


Multiply the minimum exponent of each base and you get: 
{{{2^1}}}  x  {{{3^0}}}  x  {{{7^0}}}  x  {{{13^0}}}  = 2
*****THE GCD FOR 42,96,104,18 = 2*****



PROBLEM #1:  LCM
45: {{{2^0}}}  x  {{{3^2}}}  x  {{{5^1}}}
36: {{{2^2}}}  x  {{{3^2}}}  x  {{{5^0}}}
24: {{{2^3}}}  x  {{{3^1}}}  x  {{{5^0}}}


Next, find the MAXIMUM exponent of each base:
for 2, the MAXIMUM exponent is 3, written as {{{2^3}}} 
for 3, theMAXIMUM exponent is 2, written as  {{{3^2}}}
for 5 the MAXIMUM exponent is 1, written as  {{{5^1}}}


Multiply the MAXIMUM exponent of each base and you get: 
{{{2^3}}}  x  {{{3^2}}}  x  {{{5^1}}}  = 4x9x5=360

******THE LCM FOR 45,36,24 = 360*****




PROBLEM #2:  LCM
42:    {{{2^1}}}  x  {{{3^1}}}  x  {{{7^1}}}  x  {{{13^0}}}
96:    {{{2^5}}}  x  {{{3^1}}}  x  {{{7^0}}}  x  {{{13^0}}}
104:  {{{2^3}}}  x  {{{3^0}}}  x  {{{7^0}}}  x  {{{13^1}}}
18:    {{{2^1}}}  x  {{{3^2}}}  x  {{{7^0}}}  x  {{{13^0}}}



Next, find the MAXIMUM exponent of each base:
for 2, the MAXIMUM exponent is 5, written as {{{2^5}}} 
for 3, the MAXIMUM exponent is 2, written as {{{3^2}}}
for 7 the MAXIMUM exponent is 1, written as {{{7^1}}}
for 13 the MAXIMUM exponent is 1, written as {{{13^1}}}


Multiply the MAXIMUM exponent of each base and you get: 
{{{2^5}}}  x  {{{3^2}}}  x  {{{7^1}}}  x  {{{13^1}}}  = 26,208
*****THE LCM FOR 42,96,104,18 = 26,208*****