Got doubt in the following question of Number Systems LOD II Question Number 62 Page 46.
M is a two digit number which has the property that: the product of factorials of its digits > sum of factorials of its digits. How many values of M exist? I have got the answer to be 74 but I am not sure because no explanation for the solution has been given. and the answer according to the book is 63. Can anyone explain how?
The concept here is that sum of factorials > product of factorial only when one of the numbers is 0 or1. so 10-99 total of 90 2 digit numbers. numbers which have 0 in them - 10,20,....90=9 numbers numbers which have 1 in them- 11,12,..19,21,31,41,....91= 17 numbers Also, 22 is the only number where sum of factorials=product of factorials, hence violating the condition. therefore total numbers:- 90-9-17-1=63. Hope it helps.
Find the sum of divisors of 544 excluding 1 and 544
No need to remember any formula- 544=2^5*17 now write all the combinations of the factors that can be possible, ie. 1,2,4,8,16,32 for 2^5 and 1 and 17 for 17 now if every factor is included then the sum of factors:- (2^0+2^1+2^2+2^3+2^4+2^5)(17^0+17^1) =63*18=1134 now subtract 544+1 from it to get 589. The formula given in the previous post is derived from here only. For those who find learning formulas cumbersome, the above technique will help.
Thanks man....but got others problems ryt except this one...
50^56^52 % 11 == ??
euler no. = 10 so, 56^52 % 10 = 6^52 = 6
therefore 50^6 % 11 = 6^6 = 6... but ans is given five....
The problem is in your last concept- 6^6 won't give 6 as remainder when divided by 11. by euler's theorem, 6^10 will give remainder 1 with 11. so 6^5 will give remainder -1 with 11 6^6 will give -6 or 5 as remainder. Hope it helps.
A person shooting a target,which is 465m away,hears the bullet strike it 4.5 secs after he has fired.A spectator,equidistant from the target and the shooting point(but out of line of danger),hears the bullet strike 3sec after he heard the sound of the gun.Find the speed of the sound.
a)300m/sec b)333.33m/sec c)310m/sec pls show the detailed sol.
A person shooting a target,which is 465m away,hears the bullet strike it 4.5 secs after he has fired.A spectator,equidistant from the target and the shooting point(but out of line of danger),hears the bullet strike 3sec after he heard the sound of the gun.Find the speed of the sound.
a)300m/sec b)333.33m/sec c)310m/sec pls show the detailed sol.
A person shooting a target,which is 465m away,hears the bullet strike it 4.5 secs after he has fired.A spectator,equidistant from the target and the shooting point(but out of line of danger),
dude kindly wait for the response...double post won't help....and also it is against forum rules... this is not the 1st time..please refrain yourself from such misbehavior
A person shooting a target,which is 465m away,hears the bullet strike it 4.5 secs after he has fired.A spectator,equidistant from the target and the shooting point(but out of line of danger),hears the bullet strike 3sec after he heard the sound of the gun.Find the speed of the sound.
a)300m/sec b)333.33m/sec c)310m/sec pls show the detailed sol.
Let the speed of bullet be x m/s speed of sound be s m/s Distance of the observer from shooter and target be D m each
Now, considering the 1st case, If the bullet is fired at time instant T, then the bullet will hit the target at time instant T + 465/x and the sound would be heard at time instant T + 465/x + 465/s Now the time difference between the shot fire and hearing the sound of bullet hitting the target is 4.5s i.e. (T + 465/x + 465/s) - T = 4.5 => 465/x + 465/s = 4.5
Consider the 2nd case. Again assuming that the bullet was fired at time instant T, The observer will hear the sound of shot fired at time instant T + D/s The bullet will hit the target at time instant T + 465/x The observer will hear the sound of shot hitting the target at time instant T + 465/x + D/s Now the time difference between hearing the sound of shot fire and hearing the sound of shot hitting the target is 3s i.e. (T + 465/s + D/s) - (T + D/s) = 3 => 465/x = 3
Substitute in first equation => 3 + 465/s = 4.5 => 465/s = 1.5 => s = 310 m/s
Can anybody please help me out in finding the last two digits of the foll:
45*64*76*84*32*43*57
Thanks in advance
hi, just pay attention to last two digits of each multiplication that u perform and neglect the rest: 45*64 = xx80 so take 80 80*76=xx80 80*84=xx20 20*32=xx40 40*43=xx20 20*57=xx40 so it is 40. it may look lengthy but takes 2 minutes maximum..:)
The sum of first 4 nos. is 28. the sum of the first 8 terms of d same A.P. is 88.Find the sum of the first 16 terms.
LOD1 q.29 Iam not able to understand the concept of averages which they introduced in dere solutions..
PLEASE HEL)
just use simple AP concept in the question highlighted in the bold let the terms be a,a+d,a+2d...and so on then equ will be 4a+6d=28 8a+28d=88 solve them you will get the answer
just use simple AP concept in the question highlighted in the bold let the terms be a,a+d,a+2d...and so on then equ will be 4a+6d=28 8a+28d=88 solve them you will get the answer
i really appreciate ur help...but if u have arun sharma quant book den can u please see d solutions and if possible help me in understanding the underlying concepts which they have introduced.....
i really appreciate ur help...but if u have arun sharma quant book den can u please see d solutions and if possible help me in understanding the underlying concepts which they have introduced.....
Bhai...I will try to explain that -
See, sum of 1st 4 terms is 28, hence the average would be 28/4 = 7. Also, sum of 1st 8 terms is 88, hence the average would be 88/8 = 11.
Now, as we know by Property of AP that out of 4 terms in AP like 2,4,6,8 -- Sum of 1st and last term is equal to sum of 2nd and 3rd term. Hence, average of the AP in this case i.e., 5 will be equal to the average of 2nd and 3rd term = 4+6/2 = 5 which is equal to overall average. So, wahi apply kiya hai...Dekh, average of 2nd and 3rd term is 7 and average of 4th and 5th term (jo middle terms hongi 8 terms ki AP ki) will be 11. So, the average is increasing by 4, Hence, 8th adn 7th ka average = 11+4 = 15 and so 8th and 9th ka 15+4 = 19. :)
Ab dekh, 8th and 9th ka average = 19 and hence, average of the 16 term AP is 19, Thus, Sum = 16*19 = 304
For the first case, dekho Bhai - We have to options, either to take the coin or not - So total cases will be 2^6 and there will be one case in which none is selected so minus 1. = 2^6 - 1
Now second question main if we take coin of 3paise, or coins of 2paise and 1 paise together - Total main 3paise hi add hoga hence there will redundant cases jismein sum will be repeated so we have minus some more cases in which 3 paise is included and 1 and 2 arent. Hence, kuch cases subtract honge - Thats what the Hint is saying :)
