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The Devil's Arithmetic

by Jane Yolen

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Blokova's Loss of Fingers in The Devil's Arithmetic


In The Devil's Arithmetic, Blokova loses her fingers as a punishment for failing to maintain order in the concentration camp. Each time she fails to control the prisoners or an escape occurs, she loses a finger, symbolizing the harsh and brutal penalties imposed on those in positions of authority within the camp.

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How did the blokova lose her third finger in The Devil's Arithmetic?

Although the Nazis are in overall control of the concentration camp, they put Jewish prisoners in charge of individual blocks to make life easier for themselves. It's an onerous position of responsibility, and the block commandants must maintain absolute control over the other prisoners on pain of serious punishment.

One of these commandments—called the blokova—is clearly having great difficulty keeping the prisoners under her supervision in line. She's already lost two fingers as punishment for failing to control her prisoners, and now she's lost a third because of an unsuccessful escape attempt by Shmuel and Yitzchak. It makes no difference to the Nazis that they didn't manage to escape the camp; all that matters to them is that all the prisoners—including the blokova—are kept firmly in their place by whatever means necessary.

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How did the blokova lose her third finger in The Devil's Arithmetic?

The blokova lost her third finger because of the escape attempt led by Shmuel and Yitzchak.  Even though the attempt was a failure, the fact that it occurred at all showed that, in the twisted hierarchy of camp organization, she had lost control of the prisoners under her watch.

The blokova is a Jewess who is put in charge of other Jews in the camp by the Germans.  Her responsibility is to keep the Jews in line, and "if she loses control of her zugangi", the inmates in her care, she is punished by having a finger chopped off.  The blokova had lost her first finger awhile back, when a group of zugangi rioted, and her second finger sometime later, when six zugangi hanged themselves one night.  These incidents, like the escape attempt, were interpreted as a lack of leadership on her part by the Germans, and the consequence of such lapses in responsibility is the barbaric practice of removing one of her fingers for every lapse (Chapter 16-17).

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How did Blokova lose her two fingers in The Devil's Arithmetic?

Blokova’s fingers were cut off by the Nazis as punishment for prisoner escapes.

The Blokova has only three fingers.  She is the leader of the prisoners, but she is a prisoner herself.  She lost the other two fingers when prisoners attempted to escape or rioted.

“…If she loses control of her zugangi, she will be a two-fingered whatever –you-cal- her,” said Rivka, smiling …. (p. 137)

Blokova does lose the third finger, when Shmuel and Yitzchak try to escape.  They are not successful, but she still has to be punished to set an example.

The Nazis were able to control the prisoners by getting other prisoners to control them for them.  It is much easier to punish one person when things go wrong than to try to control all of the prisoners.

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