Okay, let's tackle this problem step by step. So, the question is about finding the smallest possible number of leaves in a book, given certain conditions. Let me start by understanding the problem properly.

First, the problem defines a "book" as being composed of "n leaves," where each leaf is a single piece of paper. But here's the catch: each leaf has two pages, one on each side. So, if a book has n leaves, the total number of pages would be 2n. That makes sense. For example, if a book has 1 leaf (which is just one piece of paper), it would have 2 pages (front and back). If it has 10 leaves, that's 20 pages, and so on.

The problem states that the number of pages is 3 more than a multiple of 7. So, mathematically, this can be expressed as:
Number of pages ≡ 3 mod 7
Which translates to:
2n ≡ 3 mod 7

Additionally, the number of leaves (n) must be greater than 100. So, we need to find the smallest integer n > 100 such that 2n ≡ 3 mod 7.

Alright, so the key here is to solve the congruence equation 2n ≡ 3 mod 7 for n, and then find the smallest n > 100.

Let me recall how to solve linear congruences. The general form is ax ≡ b mod m. To solve for x, we can find the modular inverse of a modulo m, provided that a and m are coprime. In this case, a is 2 and m is 7. Since 2 and 7 are coprime (their greatest common divisor is 1), the inverse of 2 modulo 7 exists.

The inverse of 2 mod 7 is a number x such that 2x ≡ 1 mod 7. Let's find that. Testing small numbers:
2*1=2 mod7=2≠1
2*2=4 mod7=4≠1
2*3=6 mod7=6≠1
2*4=8 mod7=1
Ah, there we go. So, 2*4=8≡1 mod7. Therefore, the inverse of 2 mod7 is 4.

So, multiplying both sides of the congruence 2n ≡3 mod7 by 4 gives:
4*2n ≡4*3 mod7
Which simplifies to:
8n ≡12 mod7

But 8 mod7 is 1, and 12 mod7 is 5. So, this reduces to:
1n ≡5 mod7
Therefore, n ≡5 mod7.

So, the solutions for n are all integers congruent to 5 modulo7. That is, n=7k +5 for some integer k.

But we need n>100. So, we need to find the smallest k such that 7k +5 >100.

Let's solve for k:
7k +5 >100
7k >95
k >95/7
Calculating 95 divided by7: 7*13=91, 95-91=4. So, 95/7=13 +4/7≈13.571. Therefore, k must be greater than 13.571, so the smallest integer k is 14.

Plugging k=14 into n=7k+5:
n=7*14 +5=98 +5=103.

Wait, but let me check if 103 is indeed greater than 100. Yes, 103>100. So, is 103 the answer? Let me verify.

But hold on, let's check if 2n ≡3 mod7 for n=103.

2*103=206. Now, 206 divided by7: 7*29=203, so 206-203=3. Therefore, 206≡3 mod7. Which is exactly what we needed. So, 2n≡3 mod7 holds for n=103.

But wait, is there a smaller n>100? Let's see. The next possible n would be 103-7=96, but 96 is less than 100, so that's not acceptable. The next one would be 103+7=110, but 103 is already the first one above 100. So, 103 is the smallest possible number of leaves greater than 100 that satisfies the condition.

But let me double-check. Let's take n=103. Then pages=206. 206 divided by7 is 29 with remainder 3. So, yes, 206=7*29 +3. Therefore, 206≡3 mod7. That's correct.

Is there any number between 100 and 103 that satisfies n≡5 mod7? Let's check.

Numbers greater than 100: 101,102,103,104,105,...

Check if any of these are ≡5 mod7.

Compute 101 mod7: 101 divided by7 is 14*7=98, remainder 3. So, 101≡3 mod7.

102: 102-98=4, so 102≡4 mod7.

103: 103-98=5, so 103≡5 mod7. Yes, that's the first one. So, 103 is indeed the smallest n>100 where n≡5 mod7.

Therefore, the answer should be 103.

But wait, let me check once again. Maybe I made a mistake in the modular inverse step.

Original congruence: 2n ≡3 mod7.

Inverse of 2 mod7 is 4, so multiplying both sides by4: 8n ≡12 mod7.

But 8≡1 mod7, so 8n≡n mod7. 12≡5 mod7. Therefore, n≡5 mod7. Correct.

So, n=7k+5. To find n>100, solve 7k+5>100 =>7k>95 =>k>13.571. So, k=14, n=7*14+5=103. Correct.

Therefore, the smallest possible number of leaves is 103.

But wait, let's check if there's a smaller n>100. For example, n=101,102,103,...103 is the first one. 101: 101-100=1. 103-100=3. So, 103 is the third number after 100. Wait, but in terms of modulus, n=5 mod7. So, 103 is indeed the first number above 100 that is 5 mod7. Let me check 100: 100 divided by7 is 14*7=98, remainder 2. So, 100≡2 mod7. Then 101≡3,102≡4,103≡5. So yes, 103 is the first number above 100 that is 5 mod7.

Therefore, the answer is 103. I think that's solid.

Wait, but let me check if there's a possibility that n could be 100. But the problem states that the number of leaves is greater than 100, so n must be at least 101. But 103 is the first n>100 that satisfies the condition. So, 103 is the answer.

Alternatively, maybe there's a mistake in the initial setup. Let me verify again.

The problem says "the number of pages is 3 more than a multiple of 7". So, pages = 7k +3 for some integer k. Since pages=2n, then 2n=7k +3. So, 2n ≡3 mod7. That's correct. So, n≡5 mod7. Therefore, n=7m +5. Then, n>100. So, 7m +5 >100. 7m >95. m>13.571. So, m=14. Then, n=7*14 +5=98 +5=103. Correct.

Yes, that all checks out. So, the answer is 103. Therefore, the smallest possible number of leaves is 103.
Therefore, the smallest possible number of leaves in the book, given the conditions, is $\boxed{103}$.
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