At the end of the Year 1 the population will be = 110,000 + 4% of 110,000 =P1At the end of the Year 2 the population will be=P1+ 4% of P1= 110,000 + 4% of 110,000 + 4% of (110,000 + 4% of 110,000) = 110,000 + 4% of (110,000+ 110,000 + 4% of 110,000) = 110,000 + 2*4% of 110,000 + (4%)2 of 110,000= P2At the end of the Year 3 the population will be= P2+ 4% of P2= 110,000 +2* 4% of 110,000 + (4%)2 of 110,000 + 4% of [ 110,000 +2* 4% of 110,000 + (4%)2 of 110,000] = 110,000 +2* 4% of 110,000 + (4%)2 of 110,000 + 4% of 110,000 + 2* (4%)2 of 110,000 + (4%)3 of 110,000 =110,000 +3* 4% of 110,000 +3* (4%)2 of 110,000 + (4%)3 of 110,000 =P3At the end of the Year 4 the population will be= P3+ 4% of P3=110,000 +3* 4% of 110,000 +3* (4%)2 of 110,000 + (4%)3 of 110,000 +4% of [110,000 +3* 4% of 110,000 +3* (4%)2 of 110,000 + (4%)3 of 110,000] =110,000 +4* 4% of 110,000 + 6*(4%)2 of 110,000+4* (4%)3 of 110,000+ (4%)4 of 110,000 Now if we substitute n for 110,000 and r for 4% yearly rate of increase, we can rewrite,Before Year 1, at Year 0, the population, P0= n = n(1+r)0At the end of the Year 1 the population , P1= n+rn= x(1+r)1At the end of the Year 2 the population , P2= n+2rn+r2n =n(1+2r+r2)=n(1+r)2At the end of the Year 3 the population , P3= n+3rn+2r2n+r3n=n(1+3r+3r2+r3)=x(1+r)3At the end of the Year 4 the population , P4= n+4rn+6r2n+4r3n+r4n=n(1+r)4. . . . . . At the end of the Year n the population , Px=n+nrx+(x-1)r2n+(x-2)r3n+...........+(x-x+2)r(x-1)x+(x-x+1)rxn=n(1+r)x.....At the end of the Year 16 the population , P16= n+16rn+15r2n+14r3n+.............+3r14n+2r15n+r16n=n(1+r)16 Thus under given condition of rate of growth, the Population P(x) at xth year will be P(x)=n(1+r)x Therefore, a population of 110,000 growing at 4% per year, in 16 years will be= 110,000(1+4/100)16= 110,000(1.04)16=206027.93702999115428132473599427≈206028
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When we divide 5655 by N, we get remainder of 11, which means that 5655-11 is a multiple of N.
5655 - 11 = 5644is a multiple of N.
Similarly, 5879-14 should be a multiple of N.
5879 - 14 = 5865 is a multiple of N.
Because 5644 and 5865 are both multiples of N, their difference must be a multiple of N.
5865 − 5644 = 221 then 221 is a multiple of N.
We have three number of which N can be a multiple of, however we choose to factorize the smallest possible number amongst these three, which is 221. (This is only for simplification of the solution, smaller the number, less the factors)
221 : 1, 13, 17, 221.
There are only two two - digit factors: 13 and 17.
We divide 5865 and 5644 by both numbers.
Looking at these results, we know only 17 divides all three numbers.