find the value of a+b+c+d if product of first 10 natural numbers is written as 2^a . 3^b . 5^c .7^d
Answers
Answered by
0
6x+py=5
6x+py-5+0
ax+by+c=0; here a=6 b=p c=-5
3x+4y=2
3x+4y-2=0
here a=3, b=4, c=-2
if they have no solution then a1/a2=b1/b2≠c1/c2
then substitute the values
we get 6/3=p/4≠-5/-2
the value of p is 8
6x+py-5+0
ax+by+c=0; here a=6 b=p c=-5
3x+4y=2
3x+4y-2=0
here a=3, b=4, c=-2
if they have no solution then a1/a2=b1/b2≠c1/c2
then substitute the values
we get 6/3=p/4≠-5/-2
the value of p is 8
ghousi:
sorry this answer is wrong i was writing someother answer bymistake wrote here plz report and delete this answer plz
report and delete this answer
sorry
Answered by
5
2^a . 3^b . 5^c . 7^d = 1*2*3*4*5*6*7*8*9*10
=1*2*3*2^2*5*2*3*7*2^3*3^2*2*5
=2^8*3^4*5^2*7^1
so a=8, b=4, c=2 and d=1
a+b+c+d=15
=1*2*3*2^2*5*2*3*7*2^3*3^2*2*5
=2^8*3^4*5^2*7^1
so a=8, b=4, c=2 and d=1
a+b+c+d=15
Similar questions