If P represents a single digit in the four digit number x=17P6, how ma

Question

If P represents a single digit in the four digit number x = 17P6, how many different values of P are there such that x is divisible by 8?

A. 0
B. 1
C. 2
D. 3
E. 4

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rajatchopra1994 · Answer

Explanation:
Divisible rule for 8: Numbers are divisible by 8 if the number formed by the last three individual digits is evenly divisible by 8.

So last 3 Digits of given no. 7P6
P can take values from 0 to 9.
But it will be divisible by 8 only when P=3,7 
total values of P are 2.

IMO-C

LevanKhukhunashvili · Answer

As I dont possess any quant superpowers I started with simple things
The number is divided by 8 if it's last 3 digits divide by 8
It's a tremendous job to test all values 0-9 in 7P6

If the number is divided by 8 it means that it is for sure divided by 4
testing last 2 digits is way easier than testing last 3

06 - NO
16 - Yes
26 - No
36 - yes
46 - No
56 - yes
66 - No
76 - yes
86 - No
96 - Yes

We have excluded several options and are left with 1, 3, 5, 7, 9
By the way if we take a careful look at the testing process we can observe some Yes-No cyclicity (That came to my mind)

Then I started testing routine division by 8

716 - No
736 - yes
And of course, the cyclicity is repeating
Count carefully, only 736 and 776 satisfies my newly born theory

Pressed C and the green light went on

P.S Of course this is practice and I can afford to rely on such suspicious observations. On real test I would have tested all five options, avoiding any risk

Happy learning

Pressed C and the green light went on P.S Of course this is practice and I can afford to rely on such suspicious observations. On real test I would have tested all five options, avoiding any risk Happy learning

shiva1325 · Answer

given 17p6
we know that for the no. to be divisible by 8 last three digits must be divisible by 8
consider last 3 digits = 7p6=>this can be written as  700+p*10+6  
=>706+10*p
now find the remainder when 706+10*p is divided by 8
we get 2+2p i.e 706 divided by 8 gives remainder 2 and 10p divided by 8 gives 2p
=>2(1+p) for the no. to be divisible by 8 p+1 must be a multiple of 4 
=>p=3,p=7
hence choice  C

ScottTargetTestPrep · Answer

Recall that the divisibility rule for 8 is that the last three digits of the number must form a number that is divisible by 8. Therefore, we need to focus only on the number 7P6. If 7P6 is divisible by 8, the quotient of 7P6 divided by 8 must be a 2-digit number, and the units digit (of the 2-digit number) must be either 2 or 7 since 8 x 2 = 16 and 8 x 7 = 56. Furthermore, since a number in the 700s, when divided by 8, yields a quotient in the high 80s in the low end and a quotient in the high 90s in the high end, the only candidates for the quotient are 87, 92 and 97. Since 87 x 8 = 696, 92 x 8 =  736  and 97 x 8 =  776 , we see that P could be 3 or 7.

Answer: C

metalhead2593 · Answer

I actually forgot the divisibility rule on this one, yet I got another approach:

17P6 = 1700 + 10P + 6 = (8*212+4) + 10P + 6 = 8*213 + 2 + 10P.

So in order for 17P6 to be divisible by 8, we have to choose P such that 10P leave a remainder of 6 when divided by 8.

10P can be from 10 to 90
So 10P - 6 will be (4, 14, 24, 34, 44, 54, 64, 74, 84, 94)

Only 24 and 64 fits our criteria, hence 10P is 30 and 70. P is {3; 7}

blawp · Answer

I quickly long divided here starting with 1706 until I got to a valid number at 1736. Then, I knew that 1776 would be divisible, because you only get 6 as the units digit from 1736 if you add 8*5=40. From there, that was sufficient for me to press C.

Tushar12! · Answer

If we go through the question If p can be replaced by 1 ; 5 ; 9 still it is divisible by 8 so and can be 3 as well some one can help if I am going wrong

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Bunuel · Answer

How are you getting this? Neither 1716, 1756, nor 1796 is divisible by 8. P can take only two values: 3 and 7. Only 1736 and 1776 are divisible by 8.

If P represents a single digit in the four digit number x=17P6, how ma

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If P represents a single digit in the four digit number x=17P6, how ma

Prep Toolkit

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