1. Please explain how (60003)^4 is 6^4 followed by 16 ‘0’s (I didn’t get the 16 ‘0’s part) 😦
2. Say, if the numbers had been of the form [8m+3] & [9n+2]; then, we would have proceeded in the traditional approach?
3. Perfect symmetry in equilateral triangles is the assumption, I reckon. This is not valid for any other triangle (such as a scalene triangle), is it?
1. 60000^4 = 6^4 * 10000^4 = 6^4 followed by 16 zeroes na? the remaining terms in the binomial expansion are much smaller and hence the number of digits will be determined only by this!
2. Yes we would. You should be ready to do things the hard way if need be, but you should also be ready to seize an opportunity to do something more easily, as here.
3. Yes, but by the same token it would not be asked for another type of triangle 🙂
shouldn’t it be 20th digit from RIGHT i.e. from last when we come back and approach towards start? and1 will be1 st digit from left then..
No, Please think. You’re missing the whole point of the question. We WANT the 1st digit from right, but the question is asking it is a roundabout manner.
Hi sir, in the third answer, you have written, the values of 2 perpendiculars would be 0, i am not able to understand why though, could you please explain this
CAT-holics 
1. Please explain how (60003)^4 is 6^4 followed by 16 ‘0’s (I didn’t get the 16 ‘0’s part) 😦
2. Say, if the numbers had been of the form [8m+3] & [9n+2]; then, we would have proceeded in the traditional approach?
3. Perfect symmetry in equilateral triangles is the assumption, I reckon. This is not valid for any other triangle (such as a scalene triangle), is it?
Thanks 🙂
1. 60000^4 = 6^4 * 10000^4 = 6^4 followed by 16 zeroes na? the remaining terms in the binomial expansion are much smaller and hence the number of digits will be determined only by this!
2. Yes we would. You should be ready to do things the hard way if need be, but you should also be ready to seize an opportunity to do something more easily, as here.
3. Yes, but by the same token it would not be asked for another type of triangle 🙂
regards
J
shouldn’t it be 20th digit from RIGHT i.e. from last when we come back and approach towards start? and1 will be1 st digit from left then..
No, Please think. You’re missing the whole point of the question. We WANT the 1st digit from right, but the question is asking it is a roundabout manner.
regards
J
didn’t get it..:(
Then leave it.
Hey,
In Q 2 can we do something like
remainder is (-4) for 7
(-3) for 8
(-2) for 9…
(0) for 11 and therefore (1) for 12?
Afraid not, no guarantee that the pattern will hold.
regards
J
Hi sir, in the third answer, you have written, the values of 2 perpendiculars would be 0, i am not able to understand why though, could you please explain this
You’re dropping perpendiculars from one of the vertices. Draw and see for yourself.
regards
J