the answer is b. 177(base 10)
even i am not getting this answer.
wat is d source of ur question? puyss pls try solving this question
wat is d source of ur question? puyss pls try solving this question
If (2146!)in base 10 = (x)in base 26, then what will be the number of consecutive zeroes at the end of x?
a. (85)base 10 b. (177)base 10 c. (534)base 10 d. (17base 10 e. None of these
base 10 e. None of these
answer can be b) 177 or d)178 .. i dunno how to find which is the appropriate answer...
if options are close its better to leave the problem rather than attending it
answer can be b) 177 or d)178 .. i dunno how to find which is the appropriate answer...
if options are close its better to leave the problem rather than attending it
but how are u arriving at 177 or 178??
dude i guess we need to find for seven digits...
so i have used x^5 for the second last digit and the multiplied it to the rest of the eqn....wats the overall...and if i m wrong can u solve it for me using the same concept:) thnxx
Of course it is seven digit number but the second last digit is already decide, which is 5, and its position is also fixed. So, the remaining digits are 6. If the sum of all seven digits is 9 then sum of remaining 6 digits is 4.
hence, x1+x2+x3+x4+x5+x7 = 4. Now, x1,x2,x3,x4,x5,x7>=0.
Hence find out coeff on x^4 in (1+x+x^2+x^3+....+x^9)^6
= (1-x^10)(1-x)^-6
=(1-x^10)*(1+6C1*x + 7C2*x^2+8C3*x^3+9C4*x^4)
= 9C4. Total numbers will be 6!*9C4
If x1 = 0 then the number will become a 6 digit number.
x2+x3+x4+x5+x7 = 4. Now, x2,x3,x4,x5,x7>=0
Hence find out coeff on x^4 in (1+x+x^2+x^3+....+x^9)^5
=(1-x^10)*(1+5C1*x + 6C2*x^2+7C3*x^3+8C4*x^4)
= 8C4. Total numbers will be 5!*8C4
So, finally total is = 6!*9C4 - 5!*8C4
Correct me if I am wrong.
Of course it is seven digit number but the second last digit is already decide, which is 5, and its position is also fixed. So, the remaining digits are 6. If the sum of all seven digits is 9 then sum of remaining 6 digits is 4.
hence, x1+x2+x3+x4+x5+x7 = 4. Now, x1,x2,x3,x4,x5,x7>=0.
Hence find out coeff on x^4 in (1+x+x^2+x^3+....+x^9)^6
= (1-x^10)(1-x)^-6
=(1-x^10)*(1+6C1*x + 7C2*x^2+8C3*x^3+9C4*x^4)
= 9C4. Total numbers will be 6!*9C4
If x1 = 0 then the number will become a 6 digit number.
x2+x3+x4+x5+x7 = 4. Now, x2,x3,x4,x5,x7>=0
Hence find out coeff on x^4 in (1+x+x^2+x^3+....+x^9)^5
=(1-x^10)*(1+5C1*x + 6C2*x^2+7C3*x^3+8C4*x^4)
= 8C4. Total numbers will be 5!*8C4
So, finally total is = 6!*9C4 - 5!*8C4
Correct me if I am wrong.
shudnt it be (1-x^10)^6(1-x)^-6???
and didnt get the expansion u have done...pls elucidate...
What is the answer for this question!!!!
1: Find the remainder when 8^643 is divided by 132
shanks4mba Saysbut how are u arriving at 177 or 178??
number of zeros at the end of (2146!) in base 10 is 534
number of zero for 10! is 2 while that of 26! is 5 .. so we have a difference of 3
so number of zeros at the end = 534/3 = 178.. but one cant rule out 177 here .. its definitely a possibility..so i am not sure whether its 177 or 178
ps : obtained this method by sampling it to smaller power like octal to decimal
What is the answer for this question!!!!
1: Find the remainder when 8^643 is divided by 132
8^643 = 2^1929
132 = 2^2*33
so 2^1929mod (2^2*33) = 2^1927mod33
euler's no of 33 = 20
so 2^7mod33 = 128mod33 =29
so remainder = 29*4 = 116
shudnt it be (1-x^10)^6(1-x)^-6???
and didnt get the expansion u have done...pls elucidate...
Yes it will be.Just missed it while doing but that won't affect the answer
(1-x)^-1 = 1+x+x^2+x^3+....
= 1+x+x^2+x^3+...+x^9+x^10(1+x+x^2+x^3+....)
=>(1-x^10)(1-x)^-1 = 1+x+x^2+x^3+...+x^9
=>1+x+x^2+x^3+...+x^9 = (1-x^10)^6*(1-x)^-6
Hope I am clear.
Yes it will be.Just missed it while doing but that won't affect the answer
(1-x)^-1 = 1+x+x^2+x^3+....
= 1+x+x^2+x^3+...+x^9+x^10(1+x+x^2+x^3+....)
=>(1-x^10)(1-x)^-1 = 1+x+x^2+x^3+...+x^9
=>1+x+x^2+x^3+...+x^9 = (1-x^10)^6*(1-x)^-6
Hope I am clear.
i got it till here...how did u get this(inbold parts)
=(1-x^10)*(1+6C1*x + 7C2*x^2+8C3*x^3+9C4*x^4)
i got it till here...how did u get this(inbold parts)
=(1-x^10)*(1+6C1*x + 7C2*x^2+8C3*x^3+9C4*x^4)
(1-x)^-n = 1+nC1*x+(n+1)C2*x^2+....and so on. Its just the expansion
What is the answer for this question!!!!
1: Find the remainder when 8^643 is divided by 132
answer is 29*4 = 116 ... apply euler number concept and simplify
jain_ashu Says(1-x)^-n = 1+nC1*x+(n+1)C2*x^2+....and so on. Its just the expansion
are u doing binomial expansion?
in tht case it shud be
(a + b)^n = a^n + nC1 a^n-1b + nC2 a^n-2b^2 + ... + b^n
rt?
What is the answer for this question!!!!
1: Find the remainder when 8^643 is divided by 132
euler of 132 = 132(1-1/2)(1-1/3)(1-1/11) = 20
so 8^643 = 8^640+3
so 8^3 mod 132 = 116.....wats the answer ?????
euler of 132 = 132(1-1/2)(1-1/61) = 60
so 8^643 = 8^636+7
so 8^7 mod 132 = 68.....wats the answer ?????
132 = 2 * 66 and not 2 * 61
Originally Posted by msabhi View Post
How many 7-digit numbers exist which are divisible by 9 and whose last but one digit is 5?
What is the eqn use have used? I think it is x1+x2+x3+x4+x5+x6 = 4.
In that case it will be coeff of x^4 in (1+x+x^2+...+x^9)^6
= (1-x^10)(1-x)^-6 = (1-x^10)*(1+6C1*x+7C2*x^2+8C3*x^3+9C4*x^4+...)
=9C4 = 126.
But this contains the condition where x1 = 0.
So, use coeff of x^4 in (x+x^2+x^3+...+x^9)*(1+x+x^2+...+x^9)^5 and then find out
Puys I'm taking a straight route,
nos between 1000000 ~ 9999999 divisible by 9 --> (9999999 - 999999) / 9 = 1000000 (9s)
note: 999999 used for calculation convenience, hence 1 is not added to the result.
In every 100, we have 11, 9s out of which 1 will have 5 @ 10's digit (ex 54, 53, 52 etc)
Hence, every 11 out of 1000000(9s) has 1 multiple of 9 with 5 @ 10's digit
every 1100 ---------------- has 101 multiples of 9 --------------
every 110000-------------- has 10101 multiples of 9-------------
110000 has 10101(9s)
1000000 has 10101*1000000/110000 = 10101 * 9.09 = 91818
pls check and let me know, if you have any observation..
naga25french Says132 = 2 * 66 and not 2 * 61
already corrected my mistake..was carried away.....
are u doing binomial expansion?
in tht case it shud be
(a + b)^n = a^n + nC1 a^n-1b + nC2 a^n-2b^2 + ... + b^n
rt?
I am talking about negative power. Your expansion is valid only for positive powers.
jain_ashu SaysI am talking about negative power. Your expansion is valid only for positive powers.
hey do u have a gud source on expansions...?
puys this was a very old post, we didnt find the answer yet, pls try
Originally Posted by v!vek@rora View Post
Divisors & sum of divisors of 12^33+34^23+2^47
SK
Puys,solve this question...
Nobody is trying to do this question....!