# The gardener picked 30 green and 30 red apples. He put them in several baskets in such a way that in all baskets there were equally red apples, but different the number of greens (i.e. there were no two baskets in which there would be equally green apples). What is the largest number of baskets he could have?​

The gardener picked 30 green and 30 red apples. He put them in several baskets in such a way that in all baskets there were equally red apples, but different the number of greens (i.e. there were no two baskets in which there would be equally green apples). What is the largest number of baskets he could have?​
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allelog5
red - 30 apples, equally in each basket;
green ---- 30 apples, miscellaneous in baskets
the largest number of baskets -- ? box
1). To maximize the number of baskets with green apples, it should be assumed that the first one has 1 apple, and each next one has one more than the previous one.
We have an arithmetic sequence, 1; 2; 3; ... ; n
the sum of which is 30. For this sequence a₁ = 1; d=1, so the number of apples in the last basket is the same as the maximum number of baskets.
$S=\left(2{а}_{1}+d\left(n-1\right)\right)n/2=30$
(2*1 + 1*(n - 1))n = 30*2
(2 + n - 1)n = 60
${n}^{2}+n-60=0$
$n=\left(-1±\sqrt{1+4\ast 60}\right)/2$
${n}_{1}=\left(-1+\sqrt{241}\right)/2\approx \left(-1+15.5\right)/2\approx 7.3$ (short)
${n}_{2}=\left(-1-\sqrt{241}\right)/2\approx \left(-1-15.5\right)/2\approx -8.3$ (short) doesn't make sense since the number of baskets is positive.
This means that 30 green apples, subject to the condition, can be placed in a maximum of 7 baskets.
2). Since by condition there are the same number of red apples in each basket, the number of baskets must be a divisor of 30.
Divisors of 30: 1; 2; 3; 5; 6; ten; fifteen; thirty.
Since green apples can be placed in a maximum of 7 baskets, and 7 is not a divisor of 30, the maximum possible number is 6. Then 30 : 6 = 5 (red apple) will be in each basket.