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\(\frac{(4^7 + 4^8 + 4^9 + 4^10)}{5 }\)

Take out common factor of 4^7:
\(\frac{ 4^7 (1+ 4 + 4^2 + 4^3)}{5 }\)
=\(\frac{ 4^7 (1+ 4 + 16 + 64)}{5 }\)
=\(\frac{ 4^7 (85)}{5 }\)

=\( 4^7 (17)\)

Answer E
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If \(\frac{4^7 + 4^8 + 4^9 + 4^{10}}{5}\) is x times 4^7, what is the value of x ?

A. 1/5
B. 4/5
C. 5
D. 13
E. 17

4^7(1+4+16+64)/5
=>4^7(85)/5
=>4^7 * 17
Therefore IMO E
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On taking common we get 4^7*k/5.
k will be (1+4+16+64).this is greater than a, b, and c.Now we have two options left 13 and 17.
On solving further we get (4^7 *85)/5 which is 17.
Thus E.

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sj296
\(\frac{(4^7+4^8+4^9+4^10)}{5}\) = \(x*4^7\)

After solving =\( 17*4^7 = x*4^7 \)

=> x = 17

Option E

Hi, Bunuel I am not able to understand how we got seventeen on the left side of the equation. For 4^7+ 4^8 + 4^9 + 4^10 I used the unit digit method, by which I added the last digits of the pattern of 4 (4^7 = 4) and hence added 4+6+4+6 which gives us a "Zero in the end". Can you explain the process and where I made the mistake?

Thanks Bunuel
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Bunuel
raghav1708
sj296
\(\frac{(4^7+4^8+4^9+4^10)}{5}\) = \(x*4^7\)

After solving =\( 17*4^7 = x*4^7 \)

=> x = 17

Option E

Hi, Bunuel I am not able to understand how we got seventeen on the left side of the equation. For 4^7+ 4^8 + 4^9 + 4^10 I used the unit digit method, by which I added the last digits of the pattern of 4 (4^7 = 4) and hence added 4+6+4+6 which gives us a "Zero in the end". Can you explain the process and where I made the mistake?

Thanks Bunuel

Yes, the units digit of 4^7 + 4^8 + 4^9 + 4^10 is 0 but how is this helpful?

If \(\frac{4^7 + 4^8 + 4^9 + 4^{10}}{5}\) is x times 4^7, what is the value of x ?

A. 1/5
B. 4/5
C. 5
D. 13
E. 17

Given:

    \(4^7*x=\frac{4^7 + 4^8 + 4^9 + 4^{10}}{5}\)

Factor out 4^7 from the numerator:

    \(4^7*x=\frac{4^7*(1 + 4 + 4^2 + 4^3)}{5}\)

Reduce the expression by 4^7:

    \(x=\frac{1 + 4 + 4^2 + 4^3}{5}\);

    \(x=\frac{85}{5}=17\).

Answer: E.


Got it! Thank you so much Bunuel
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