Točno
1. lipnja 2015. 00:45 (9 godine, 1 mjesec)
Upozorenje: Ovaj zadatak još niste riješili!
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Uvidimo da broj znamenaka nekog prirodnog broja
u nekom brojevnom sustavu baze
je ![\lfloor\log _b a\rfloor+1](/media/m/d/e/5/de526bfd51ad4f130f0fec3352fd4089.png)
gdje je
cijeli broj a
decimalni ostatak u intervalu ![[0, 1\rangle](/media/m/9/7/e/97ef2911bb7e97db0c25928d50437ec1.png)
vidljivo je da![\lfloor\log _b a\rfloor = \log _ba - x](/media/m/b/4/1/b414640491e0a344178349d29a801157.png)
![= \lfloor \log _{10} 2^{1997}\rfloor + 1 + \lfloor \log _{10} 5^{1997}\rfloor + 1](/media/m/0/9/8/09872f62876868123c8eedae063ed24a.png)
![= \log _{10}2^{1997} - x_1 + \log _{10}5^{1997} - x_2 +2](/media/m/e/9/c/e9c79503604fe527433fae8d568c7244.png)
![= \log _{10}10^{1997} - (x_1 + x_2) +2](/media/m/6/6/7/667e4a8de73bb921860ae06ffc52c370.png)
Budući da je zbroj znamenaka prirodan broj, a
i
cijeli brojevi,
mora biti cijeli broj. Budući da
,
može biti
ili
.
Budući da
i
nisu potencije broja
,
pa
može biti samo
.
![\log _{10}10^{1997} - 1 + 2 = 1997 + 1= 1998](/media/m/2/1/5/21514ea3c79531621ae9563cee9d4953.png)
![a](/media/m/6/d/2/6d2832265560bb67cf117009608524f6.png)
![b>1](/media/m/4/9/4/49478db619dc58f5655fe2c6277612e8.png)
![\lfloor\log _b a\rfloor+1](/media/m/d/e/5/de526bfd51ad4f130f0fec3352fd4089.png)
![\log _b a = k + x](/media/m/f/2/6/f261d2529c4d0f6c747d0b88bd31567b.png)
![k](/media/m/f/1/3/f135be660b73381aa6bec048f0f79afc.png)
![x](/media/m/f/1/8/f185adeed9bd346bc960bca0147d7aae.png)
![[0, 1\rangle](/media/m/9/7/e/97ef2911bb7e97db0c25928d50437ec1.png)
vidljivo je da
![\lfloor\log _b a\rfloor = \log _ba - x](/media/m/b/4/1/b414640491e0a344178349d29a801157.png)
![m + n=](/media/m/2/0/1/201fe49a5bfe6ebad11cfe3f875e06ea.png)
![= \lfloor \log _{10} 2^{1997}\rfloor + 1 + \lfloor \log _{10} 5^{1997}\rfloor + 1](/media/m/0/9/8/09872f62876868123c8eedae063ed24a.png)
![= \log _{10}2^{1997} - x_1 + \log _{10}5^{1997} - x_2 +2](/media/m/e/9/c/e9c79503604fe527433fae8d568c7244.png)
![= \log _{10}10^{1997} - (x_1 + x_2) +2](/media/m/6/6/7/667e4a8de73bb921860ae06ffc52c370.png)
Budući da je zbroj znamenaka prirodan broj, a
![\log _{10}10^{1997}](/media/m/6/e/2/6e2ac5019204091ba318c3a308d11dab.png)
![2](/media/m/e/e/e/eeef773d19a3b3f7bdf4c64f501e0291.png)
![x_1+x_2](/media/m/c/b/9/cb940d9925f572fd061d61476765f566.png)
![x_1, x_2 < 1](/media/m/d/f/2/df2ba6a8809b572239334ba865674807.png)
![x_1 + x_2](/media/m/1/a/9/1a926152501fef8652568fd086480796.png)
![1](/media/m/a/9/1/a913f49384c0227c8ea296a725bfc987.png)
![0](/media/m/7/b/8/7b8b0b058cf5852d38ded7a42d6292f5.png)
Budući da
![2^{1997}](/media/m/6/9/7/697099b3905b96a9d3bc7a34f48aeadb.png)
![5^{1997}](/media/m/9/7/2/972744ce0891793b035a75317af3562a.png)
![10](/media/m/5/b/e/5beb46430dbe2d22c0f8289c36a92c84.png)
![x_1, x_2 > 0](/media/m/8/9/a/89ae2d821b6237e692010cfbe9e1b3fe.png)
![x_1 + x_2](/media/m/1/a/9/1a926152501fef8652568fd086480796.png)
![1](/media/m/a/9/1/a913f49384c0227c8ea296a725bfc987.png)
![\log _{10}10^{1997} - 1 + 2 = 1997 + 1= 1998](/media/m/2/1/5/21514ea3c79531621ae9563cee9d4953.png)
![m + n = 1998](/media/m/d/f/e/dfe74093a77616a5c28f3dc65557f155.png)