MHB Find Integer Part of A: Math Problem

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Integer
Click For Summary
The problem involves calculating the integer part of A, defined by the equation A=(16×72+17×73+18×74+19×75)/(16×71+17×72+18×73+19×74) multiplied by 150. The solution reveals that A can be expressed as 152 plus a fraction involving weighted sums of integers. The numerator of the fraction is 16×8 + 17×6 + 18×4 + 19×2, while the denominator consists of the same weights applied to the integers 71, 72, 73, and 74. This approach simplifies the calculation and highlights the cleverness of the solution. The integer part of A is thus determined to be 152.
Albert1
Messages
1,221
Reaction score
0
$A=(\dfrac{16\times72+17\times73+18\times74+19\times75}{16\times71+17\times72+18\times73+19\times74})\times 150$
find the integer part of A
 
Mathematics news on Phys.org
A = $( 1+\frac{16+ 17+ 18 + 19}{16 * 71 + 17 * 72 + 18 * 73 + 19 * 74}) * 150$
= $150 +\frac{16 * 150 + 17 * 150 + 18 * 150 + 19 *150}{16 * 71 + 17 * 72 + 18 * 73 + 19 * 74}$
= $150 + 2 +\frac{16 *8 + 17 * 6 + 18 * 4 + 19 *2}{16 * 71 + 17 * 72 + 18 * 73 + 19 * 74}$

so ans = 152
 
Last edited:
kaliprasad said:
A = $( 1+\frac{16+ 17+ 18 + 19}{16 * 71 + 17 * 72 + 18 * 73 + 19 * 74}) * 150$
= $150 +\frac{16 * 150 + 17 * 150 + 18 * 150 + 19 *150}{16 * 71 + 17 * 72 + 18 * 73 + 19 * 74}$
= $152 +\frac{16 * 8 + 17 * 6 + 18 * 4 + 19 *8}{16 * 71 + 17 * 72 + 18 * 73 + 19 * 74}$


so ans = 152
it should be:
= $152 +\frac{16 * 8 + 17 * 6 + 18 * 4 + 19 *2}{16 * 71 + 17 * 72 + 18 * 73 + 19 * 74}$indeed very smart solution:)
 

Similar threads

Replies
7
Views
2K
Replies
4
Views
1K
Replies
1
Views
1K
Replies
4
Views
3K
Replies
1
Views
1K
Replies
2
Views
2K
Replies
4
Views
2K
Replies
1
Views
1K
Replies
3
Views
2K