If #3^p+3^4=90#, #2^r+44=76#, and #5^3+6^s=1421#, what is the product of # p, r#, and #s#?

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Answer 1 · 2018-03-25T16:58:20

Given #3^p+3^4=90#, #2^r+44=76#, and #5^3+6^s=1421#,

So #3^p+3^4=90#

#=>3^p=90-81=3^2#

#=>p=2#

Again

#2^r+44=76#

#=>2^r=76-44=32=2^5#

#=>r=5#

Again

#5^3+6^s=1421#

#=>6^s=1421-125=1296=6^4#

#=>s=4#

Hence #pxxrxxs=2xx5xx4=40#

Answer 2 · 2018-03-25T16:59:06

#prs=40#

As #3^p+3^4=90#, we have #3^p=90-3^4=90-81=9#

i.e. #3^p=3^2# and hence #p=2#

Similarly #2^r+44=76# implies #2^r=76-44=32=2^5# and

hence #r=5#

and #5^3+6^s=1421# implies #6^s=1421-5^3=1421-125=1296=6^4#

and therefore #s=4#

hence #prs=2*5*4=40#