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Let a1, a2, ... , a52 be positive integers such that a1 ＜ a2 ＜ ...

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Joined: 03 Jul 2019
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Let a1, a2, ... , a52 be positive integers such that a1 ＜ a2 ＜ ...  [#permalink]

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23 Aug 2019, 10:45
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27% (02:27) correct 73% (03:16) wrong based on 11 sessions

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Let a1, a2, ... , a52 be positive integers such that a1 ＜ a2 ＜ ... ＜ a52. Suppose, their arithmetic mean is one less than the arithmetic mean of a2, a3, ..., a52. If a52 = 100, then the largest possible value of a1 is

A. 48
B. 20
C. 45
D. 23
E. 43
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Joined: 19 Oct 2018
Posts: 852
Location: India
Let a1, a2, ... , a52 be positive integers such that a1 ＜ a2 ＜ ...  [#permalink]

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23 Aug 2019, 15:29
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1
$$\frac{a_1+a_2+ ... + a_{52}}{52} = \frac{a_2+a_3 ... a_{51}+ a_{52}}{51} - 1$$

$$a_1+51*52= a_2+a_3 ... a_{51}+ a_{52}$$

We want maximize $$a_1$$, hence we have to maximize the sum of $$a_2$$, $$a_3$$, ..., $$a_{52}$$.

Also, we know that $$a_1$$ ＜ $$a_2$$ ＜ ... ＜ $$a_{52}$$, and $$a_2$$, ... , $$a_{52}$$ are positive integers. Hence we can assign the consecutive integers from $$a_2$$ to $$a_{52}$$.

$$51a_1+51*52= 50+51+.....+99+100$$

$$51a_1+51*52= \frac{51}{2}*(50+100)$$

$$51a_1=51*75-51*52$$

$$51a_1=51(75-52)$$

$$a_1=23$$

AbdulMalikVT wrote:
Let a1, a2, ... , a52 be positive integers such that a1 ＜ a2 ＜ ... ＜ a52. Suppose, their arithmetic mean is one less than the arithmetic mean of a2, a3, ..., a52. If a52 = 100, then the largest possible value of a1 is

A. 48
B. 20
C. 45
D. 23
E. 43
VP
Joined: 03 Jun 2019
Posts: 1496
Location: India
Let a1, a2, ... , a52 be positive integers such that a1 ＜ a2 ＜ ...  [#permalink]

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23 Aug 2019, 23:10
1
AbdulMalikVT wrote:
Let a1, a2, ... , a52 be positive integers such that a1 ＜ a2 ＜ ... ＜ a52. Suppose, their arithmetic mean is one less than the arithmetic mean of a2, a3, ..., a52. If a52 = 100, then the largest possible value of a1 is

A. 48
B. 20
C. 45
D. 23
E. 43

Given:
1. Let a1, a2, ... , a52 be positive integers such that a1 ＜ a2 ＜ ... ＜ a52.
2. Suppose, their arithmetic mean is one less than the arithmetic mean of a2, a3, ..., a52.
3. a52 = 100

Asked: The largest possible value of a1 is

$$\frac{a1+a2+.... + a52}{52} = \frac{a2 + a3 + .... + a52}{51} -1$$
a52 = 100
$$\frac{a1}{52} + \frac{a2 + a3 + .... + 100}{52} = \frac{a2 + a3 + .... + 100}{51} -1$$
$$\frac{a1}{52} = (a2 + a3 + .... + 100) (\frac{1}{51} - \frac{1}{52}) -1$$
$$a1 = \frac{a2+a3+...+100}{51} - 52$$
For largest possible value of a1
a2=a3-1
a3=a4-1

a51=a52-1=99

{a52,a51,.... a2} = {100,99, ... 50}

Largest possible value of a1 = $$\frac{51}{2}\frac{(100+50)}{51} -52 = \frac{51*75}{51} - 52 = 23$$

IMO D
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Let a1, a2, ... , a52 be positive integers such that a1 ＜ a2 ＜ ...   [#permalink] 23 Aug 2019, 23:10
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