Maximum value of n such that 42x57x92x91x52x62x63x64x65x66x67 is perfectly divisible by 42^n??
Last digit (36472)^123! * (34767)^76!
@anudsweet said:Maximum value of n such that 42x57x92x91x52x62x63x64x65x66x67 is perfectly divisible by 42^n??
@anudsweet said: Maximum value of n such that 42x57x92x91x52x62x63x64x65x66x67 is perfectly divisible by 42^n??
is the answewr 3 !
as the max power of 7 is 3 and 2and 3 is 5!
so minimum of two is 3!
as the max power of 7 is 3 and 2and 3 is 5!
so minimum of two is 3!
@anudsweet said: Last digit (36472)^123! * (34767)^76!
last digit of first expression : 6
last digit of Second expression : 1
so total : 6*1 = 6
@rahul005 said: Find the number of three-digit numbers that can be expressed as the sum of four consecutive,five consecutive and seven consecutive positive integers. 1)9 2)10 3)11 4)14 5)6 pls show detaied sol
Is the answer 6?
My Solution uses a method mentioned somewhere in the TIME material on numbers (somewhat lengthy though.. 
) :
) :Sum of any 4 consecutive nos starting from number a = 2(2a+3)=4a+6
Sum of any 5 consecutive nos starting from number b = 5(b+2)=5b+10
Sum of any 7 consecutive nos starting from number c = 7(c+3)=7c+21
Now, we have to find number of 3 digit numbers "x", such that
x = 4a+6 = 5b+10 = 7c+21, and 100
From the above, 7c + 21 = 5b+10 i.e. 7c+11 = 5b. Smallest solution is c=2,b=5. Putting c=2 in original expression, 7*2+21 = 35. Result: All numbers of the form 35 + Multiple of LCM(5,7) will also be of the form 7c + 21 AND 5b + 10. This general number is expressed as 35 + 35y.
Now, we use 35 + 35y = 4a+6 ie, 35y + 29 = 4a. Smallest solution is y=1,a=16. Putting y=1 in original expression, 35+35*1 = 70. Result: All numbers of the form 70 + Multiple of LCM(35,4) will also be of the form 35+35y AND 4a+6. This general number is expressed as 70 + 140z.
Hence, x is of the form 70+140z. Now 100
@rahul005 said:Find the number of three-digit numbers that can be expressed as the sum of four consecutive,five consecutive and seven consecutive positive integers. 1)9 2)10 3)11 4)14 5)6 pls show detaied sol
Let us try to find out the nature of such three-digit numbers.
Since the number can be expressed as sum of seven consecutive positive integers, it has to be a multiple of 7 ( Explanation: let a - 3, a - 2, a - 1, a, a + 1, a + 2, a + 3 be the 7 consecutive numbers, then summation of these numbers is 7a, a multiple of 7)
Similarly, the desired three digit number(s) should be a multiple of 5 and should leave a remainder of 4 when divided by 4 (4k + 2 form).
So the question boils down to: Find all three digit numbers which are divisible by 5 and 7 and leave a remainder of 2 when divided by 4. I guess it is easy here on..
So 210, 350, 490, 630, 770, 910 fulfil the crietria. Hence 6. Let me know if I am correct or not.
@RoadKill said: Is the answer 6?My Solution uses a method mentioned somewhere in the TIME material on numbers (somewhat lengthy though.. ) :Sum of any 4 consecutive nos starting from number a = 2(2a+3)=4a+6Sum of any 5 consecutive nos starting from number b = 5(b+2)=5b+10Sum of any 7 consecutive nos starting from number c = 7(c+3)=7c+21 Now, we have to find number of 3 digit numbers "x", such thatx = 4a+6 = 5b+10 = 7c+21, and 100Your logic is similar to mine... However, you can simplify it a lot. Kindly have a look at the method I have posted below.. Let me know in case you need any explanation.
P.S. Both of us shred our solutions at the same time :)
@[410661:rahul005] Quantohelp's method is better optimized than what I said earlier.
Use x = 7a = 5b = 4c+2, instead of the brute force method I gave before. The solution is arrived at quickly now.
@[592819:Quantohelp] Yeah I saw! I missed using the simplification you used. Now this question is doable without too much number crunching. 
Updated rahul005 above regarding the same.
@anudsweet said: Maximum value of n such that 42x57x92x91x52x62x63x64x65x66x67 is perfectly divisible by 42^n??
42=2*3*7
there r 13 2's,5 3's and 3 7's.
so here constraint is on 7.
hence only 3 group of (2*3*7) can be formed.
so n=3
no of zeroes in 1^1*2^2*3^3.......*150^150 ?
@Sachin27 said: no of zeroes in 1^1*2^2*3^3.......*150^150 ?
No. of zeroes : 155*15 + (25 + 50 +75 +100 +125 +150)
plz share the approach of wilson theoram in this question :
remainder of 69! / 71
@[558153:sagac]
Acc. to wilson
For a prime number p,
thus, 1
cheers :mg:
@gs4890 said: @sagac Acc. to wilsonFor a prime number p, thus, 1cheers
hey have u copied it from some material ?
can u pls share that material so that we know abt othr such theorms also!
can u pls share that material so that we know abt othr such theorms also!
@LeoN88 said: No. of zeroes : 155*15 + (25 + 50 +75 +100 +125 +150)
Shouldn't there be one more 125, to cater for three 5s in 125 ?
Final ans = 155*15 + (25+50+75+100+125+150) + (125) ??
@YouMadFellow said:Shouldn't there be one more 125, to cater for three 5s in 125 ? Final ans = 155*15 + (25+50+75+100+125+150) + (125) ??
Yes you are right..... "Silly Mistake " 
Thanks
i have some number.if i divide by 2 ,reminder 1,if i divide it by 3,also remain reminder 1,divide by 4,also reminder 1 ,divide by 5,also reminder 1,divide by 6 ,reminder 1 and finally when divide by 7 reminder 0..what is the number?